CSIR Net Physics Previous Year Question Papers

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    CSIR NET Physics Previous Year Question Papers Mock test for Students

    An open organ pipe has a frequency of 156 Hz at 15 °C. At what temperature will its frequency increase to 160 Hz?

    a) 30 °C

    b) 50 °C

    c) 70 °C

    d) 90 °C

    Explanation:
    The question explores how the frequency of sound produced by an open organ pipe changes when the temperature of the surrounding air increases. The pipe produces a known frequency at one temperature, and the task is to determine the temperature corresponding to a slightly higher frequency.

    In an organ pipe, the frequency of the sound wave depends on the speed of sound in air. The speed of sound is not constant; it increases as the temperature of air rises. This happens because warmer air molecules move faster and transfer sound vibrations more quickly. For a pipe of fixed length, the wavelength of the standing wave pattern does not change. Therefore, any change in the speed of sound directly changes the frequency of the wave.

    The speed of sound in air is proportional to the square root of the absolute temperature (measured in Kelvin). Because frequency in a pipe is proportional to the speed of sound, frequency is also proportional to the square root of the absolute temperature. To analyze the change, the temperature values must first be converted from Celsius to Kelvin. Then the ratio of the frequencies is compared with the ratio of the square roots of the corresponding temperatures. Using this proportional relationship, the unknown temperature can be determined without altering the pipe’s length.

    This relationship explains why musical instruments sound slightly higher in pitch on warmer days.

    In summary, the frequency of an organ pipe increases with temperature because the speed of sound rises with temperature, and frequency is directly related to this speed.

    Option a – 30 °C

    A pipe that is 85 cm long and closed at one end can support how many natural frequencies below 1250 Hz if the speed of sound in air is 340 m/s?

    a) 6

    b) 4

    c) 12

    d) 8

    Explanation:
    This question deals with resonance in an air column when one end of the pipe is closed and the other end is open. The goal is to determine how many natural resonant frequencies exist below a given limit.

    In a pipe closed at one end, standing waves form with a node at the closed end and an antinode at the open end. Because of this boundary condition, only odd harmonics can exist. The fundamental mode occurs when one-quarter of a wavelength fits inside the pipe length. Higher resonances occur when three-quarters, five-quarters, and other odd multiples of a quarter wavelength fit into the pipe.

    The fundamental frequency for such a pipe depends on the speed of sound and the pipe length. Using the appropriate formula for closed pipes, the lowest resonant frequency can be determined. Once this value is known, the higher resonant frequencies occur as odd multiples of this fundamental frequency. Each successive odd multiple represents another allowed standing wave pattern within the pipe.

    To determine how many frequencies exist below the given limit, these odd multiples are listed sequentially and compared with the specified maximum frequency. Only those frequencies that remain below that limit are counted.

    This behaviour is similar to instruments like clarinets, where only certain harmonic notes are naturally produced due to the closed pipe structure.

    Overall, the number of allowed resonant frequencies is determined by calculating the fundamental frequency and counting its odd multiples that fall below the specified frequency limit.

    Option a – 6

    For a pipe closed at one end with a length of 85 cm, how many natural frequencies will exist under 1250 Hz if sound travels at 340 m/s?

    a) 4

    b) 5

    c) 7

    d) 6

    Explanation:
    The problem asks how many resonant frequencies are possible in a pipe that is closed at one end and open at the other, given the pipe’s length and the speed of sound.

    When a pipe is closed at one end, the standing wave pattern must satisfy specific boundary conditions. The closed end always forms a displacement node, while the open end forms a displacement antinode. Because of this arrangement, the pipe supports only odd harmonics. The fundamental vibration occurs when one-quarter of a wavelength fits inside the pipe length.

    To begin solving the problem, the fundamental frequency must be determined using the formula that relates wave speed, pipe length, and the quarter-wavelength condition. Once the fundamental frequency is known, the next allowed frequencies occur as odd multiples such as three times, five times, and seven times the fundamental value. These correspond to different standing wave patterns that can form in the pipe.

    The next step is to list these harmonic frequencies in order and compare them with the given upper frequency limit. Only the frequencies that remain below that limit are counted. Each acceptable odd multiple represents one natural resonant frequency supported by the pipe.

    Thus, by calculating the fundamental frequency and checking its odd multiples against the given limit, the total number of natural resonant frequencies can be identified.

    Option d – 6

    Two closed organ pipes are played together, creating 5 beats per second. If their lengths are in the ratio 21:20, what are their frequencies?

    a) 100 Hz and 105 Hz

    b) 105 Hz and 100 Hz

    c) 110 Hz and 105 Hz

    d) 105 Hz and 110 Hz

    Explanation:
    This question involves the phenomenon of beats produced when two sound waves with slightly different frequencies interfere with each other.

    When two sound waves of nearly equal frequency travel together, their superposition causes Periodic increases and decreases in loudness. This effect is known as beats. The number of beats heard per second equals the absolute difference between the two frequencies. Therefore, if the beat frequency is known, the difference between the two pipe frequencies can be determined.

    In organ pipes, the frequency of the sound produced depends on the length of the pipe. For pipes of the same type and medium, the frequency is inversely proportional to the length of the pipe. This means a shorter pipe produces a higher frequency, while a longer pipe produces a lower frequency.

    Since the two pipe lengths are given in a ratio, their frequencies must follow the inverse ratio. Using this relationship, the two frequencies can be expressed in terms of a common variable. The difference between these two frequencies must equal the observed beat frequency. By forming this equation, the individual frequencies of the pipes can be calculated.

    This principle is often used by musicians to tune instruments by listening for the disappearance of beats.

    In summary, the pipe frequencies are determined by combining the inverse length–frequency relationship with the beat frequency condition.

    Option a – 100 Hz and 105 Hz

    A 340 Hz tuning fork is placed at the open end of a tube 150 cm high. Water is slowly poured in. What is the minimum water height required to achieve resonance? (Speed of sound in air = 340 m/s)

    a) 25 cm

    b) 45 cm

    c) 75 cm

    d) 95 cm

    Explanation:
    The problem describes a resonance experiment involving a tuning fork and a vertical tube partially filled with water. The goal is to determine the water level at which resonance first occurs.

    When a tuning fork vibrates near the open end of a tube, it sends sound waves into the air column inside the tube. If the length of the air column matches a natural resonant condition, a standing wave forms and the sound becomes noticeably louder. When water is poured into the tube, the length of the air column decreases.

    For a tube that behaves like a closed pipe (closed at the water surface and open at the top), the first resonance occurs when the air column length equals one-quarter of the wavelength of the sound wave. This condition allows a node to form at the closed end and an antinode at the open end.

    To analyze the situation, the wavelength of the sound must first be determined using the relation between wave speed, frequency, and wavelength. Once the wavelength is known, the quarter-wavelength condition provides the resonant air column length. The required water height can then be found by subtracting this air column length from the total tube length.

    This principle is commonly demonstrated in laboratory resonance tube experiments.

    In short, resonance occurs when the air column length becomes equal to one-quarter of the wavelength of the tuning fork’s sound wave.

    Option a – 25 cm

    A tuning fork produces 5 beats per second with an air column at 27 °C. When the temperature drops to -3 °C, beats stop. The fork’s frequency is:

    a) 70 Hz

    b) 92 Hz

    c) 104 Hz

    d) 147 Hz

    Explanation:
    This question involves both the concept of beats and the effect of temperature on the speed of sound in air.

    Beats occur when two sound waves of slightly different frequencies interfere with each other. The beat frequency equals the difference between the two frequencies. In this case, the tuning fork and the resonant air column produce beats at a certain temperature, meaning their frequencies are slightly different.

    However, when the temperature changes, the speed of sound in air also changes. Since the resonant frequency of an air column depends on the speed of sound, its frequency shifts with temperature. When the temperature drops to a new value and the beats disappear, it means the frequencies of the tuning fork and the air column have become equal.

    To analyze the problem, the relationship between the speed of sound and temperature must be considered. The speed of sound increases with temperature, and therefore the resonant frequency of the air column also increases. By comparing the resonant frequencies at the two temperatures and considering the beat frequency difference at the first temperature, the tuning fork’s frequency can be determined.

    This method is sometimes used in acoustic experiments to measure unknown frequencies.

    In summary, the tuning fork frequency can be determined by analyzing how the resonant frequency of the air column changes with temperature and how that change affects the beat frequency.

    Option b – 92 Hz

    A glass tube 1.1 m long has adjustable water levels. A tuning fork with 660 Hz is placed near the tube’s open end. How many harmonics can this tube support? (Speed of sound = 330 m/s)

    a) 2

    b) 3

    c) 4

    d) 5

    Explanation:
    This question examines resonance in a tube where the effective air column length can be adjusted using water inside the tube.

    When a tuning fork vibrates near the open end of the tube, the sound waves enter the air column. If the length of the air column matches one of the resonant conditions, standing waves form and the sound becomes louder. Because the bottom of the air column is effectively closed by the water surface, the tube behaves like a pipe closed at one end.

    In such pipes, the allowed resonant frequencies correspond only to odd harmonics. The fundamental mode occurs when the air column length equals one-quarter of the wavelength. Higher resonances occur when three-quarters, five-quarters, and other odd multiples of a quarter wavelength fit within the air column.

    The wavelength of the sound produced by the tuning fork is determined using the relationship between wave speed, frequency, and wavelength. Once the wavelength is known, the maximum possible air column length is compared with the lengths required for different harmonic modes. Each harmonic whose required length fits within the tube is counted.

    Thus, the number of harmonics supported by the tube depends on how many allowed standing-wave patterns can physically fit within the maximum air column length.

    In summary, the tube supports as many odd harmonic modes as can fit within its maximum air column length based on the wavelength of the sound.

    Option c – 4

    Light passing through a single Nicol prism becomes

    a) Unpolarised

    b) Elliptically polarised

    c) Plane polarised

    d) None of these

    Explanation:
    This question relates to the concept of polarization of Light using a Nicol prism, which is a device used to produce polarized Light.

    Light from natural sources is usually unpolarized, meaning the Electric Field vibrations occur in many different planes perpendicular to the direction of propagation. When such Light passes through certain optical devices or materials, only vibrations in a particular direction are allowed to continue.

    A Nicol prism is made from calcite crystal and is designed to separate Light into two rays using the phenomenon of double refraction. One of these rays is absorbed or diverted, while the other passes through the prism. The transmitted ray contains vibrations confined to a single plane.

    Because the prism eliminates all other vibration directions except one, the emerging Light has its Electric Field oscillating only in a specific plane perpendicular to the direction of travel. This type of Light is referred to as polarized Light.

    Such devices are widely used in optical experiments and instruments where control over the direction of Light vibration is required.

    In summary, a Nicol prism converts ordinary unpolarized light into light that vibrates in a single plane by selectively transmitting only one component of the light wave.

    Option c – Plane polarised

    Which of the following waves cannot be polarised?

    a) Sound waves

    b) Gamma rays

    c) Radio waves

    d) UV rays

    Explanation:
    The question focuses on the property of polarization and which types of waves can exhibit this behaviour.

    Polarization refers to the restriction of vibrations of a wave to a single plane. This phenomenon occurs only in transverse waves, where the direction of vibration is perpendicular to the direction of wave propagation. In such waves, the vibration direction can be selectively restricted or filtered.

    Electromagnetic waves such as visible light, radio waves, ultraviolet radiation, and gamma rays are transverse in nature. Because their electric and magnetic fields oscillate perpendicular to the direction of propagation, they can exhibit polarization.

    However, waves that are longitudinal behave differently. In longitudinal waves, the particles of the medium vibrate parallel to the direction of wave propagation. Because their vibrations occur along the direction of motion, there is no perpendicular vibration plane that can be selectively filtered.

    This is why polarization experiments are often used to confirm whether a wave is transverse.

    In summary, only transverse waves can be polarized, while longitudinal waves cannot exhibit polarization because their vibrations occur along the direction of propagation.

    Option a – Sound waves

    When a sugar solution rotates the plane of polarisation to the right, the solution is

    a) Laevo-rotatory

    b) Polarised

    c) Dextro-rotatory

    d) Unpolarised

    Explanation:
    This question relates to the optical activity of certain substances and how they affect polarized light.

    Some chemical substances have molecules that interact with polarized light in a special way. When plane-polarized light passes through such a substance, the plane in which the Electric Field vibrates may rotate. This phenomenon is known as optical rotation or optical activity.

    The direction of rotation is important when classifying optically active substances. If the plane of polarization rotates toward the right side (clockwise direction) when viewed facing the incoming light, the substance is described using a specific term. If the rotation occurs toward the left (counterclockwise), a different term is used.

    Sugar solutions are common examples of optically active substances. Their Molecular structure causes them to interact asymmetrically with polarized light, leading to a measurable rotation of the polarization plane. Instruments called polarimeters are often used to measure this rotation for chemical analysis.

    In summary, optically active substances rotate the plane of polarized light, and the direction of this rotation determines how the substance is classified.

    Option c – Dextro-rotatory

    At the polarising angle, the angle between the reflected and refracted rays is

    a) 90°

    b) 180°

    c) 45°

    d) 60°

    Explanation:
    This question concerns the special condition known as the polarising angle or Brewster’s angle. It is the angle of incidence at which reflected light becomes completely plane polarised.

    When unpolarised light strikes the boundary between two transparent media, part of it is reflected and part is refracted. At a particular angle of incidence, the reflected ray becomes completely plane polarised because the Electric Field vibrations parallel to the plane of incidence are eliminated from the reflected beam.

    At this special angle, an important geometric relationship occurs between the reflected and refracted rays. The direction of the refracted ray forms a specific angular relationship with the reflected ray. This result follows directly from Brewster’s law and the geometry of reflection and refraction at the interface.

    Mathematically, Brewster’s law relates the polarising angle to the refractive index of the medium. When this condition is satisfied, the reflected and refracted rays become mutually perpendicular. This perpendicular relationship is a key identifying feature of the polarising angle and is frequently used in Optics experiments to determine refractive indices.

    An everyday example can be seen in polarized sunglasses, which reduce glare from water or road surfaces because reflected light near Brewster’s angle becomes polarized.

    In summary, at the polarising angle the reflected light is completely plane polarised and the reflected and refracted rays form a characteristic perpendicular relationship.

    Option a – 90°

    Two Nicol prisms are SET so their principal planes are at 30°. What percentage of unpolarised light passes through?

    a) 37.5%

    b) 75%

    c) 100%

    d) None of these

    Explanation:
    This question involves the transmission of light through two polarising devices and the application of Malus’ law.

    When unpolarised light passes through the first Nicol prism, it becomes plane polarised. However, only half of the original intensity remains because the prism selects vibrations in only one direction while eliminating the rest.

    When this plane-polarised light passes through a second Nicol prism, the transmitted intensity depends on the angle between the principal planes of the two prisms. The principal plane represents the direction along which light vibrations are allowed to pass through the prism.

    Malus’ law describes how the intensity of polarized light changes after passing through a second polarizer. According to this law, the transmitted intensity is proportional to the square of the cosine of the angle between the transmission axes of the two prisms.

    To determine the final transmitted intensity relative to the original unpolarised light, two steps are considered: first the reduction caused by the first polarizer, and then the further reduction governed by Malus’ law for the second prism. By applying this relationship with the given angular separation, the percentage of transmitted light can be determined.

    In summary, the transmitted intensity depends on both the initial polarization process and the angular alignment between the two Nicol prisms as described by Malus’ law.

    Option a – 37.5%

    When light strikes a surface at its polarising angle of 57°, the correct angle of incidence is

    a) 57°

    b) 37°

    c) 90°

    d) 137°

    Explanation:
    This question is based on the concept of Brewster’s angle, also called the polarising angle. It is the angle of incidence at which reflected light becomes completely plane polarised.

    When unpolarised light falls on a transparent surface such as glass or water, it splits into a reflected and a refracted ray. Under most conditions the reflected light is only partially polarised. However, when the angle of incidence reaches a specific value, the reflected light becomes fully plane polarised.

    This special angle is known as Brewster’s angle and depends on the refractive index of the second medium. Brewster’s law states that the tangent of the polarising angle is equal to the refractive index of the medium relative to the first medium.

    At this angle, an important geometric condition also occurs: the reflected and refracted rays become perpendicular to each other. Because of this unique relationship, identifying the polarising angle directly determines the corresponding angle of incidence responsible for producing the fully polarised reflected beam.

    Such conditions commonly occur when sunlight reflects from surfaces like water, glass, or roads, which is why polarized sunglasses are designed to reduce glare produced by these reflections.

    In summary, the polarising angle directly corresponds to the angle of incidence at which reflected light becomes completely plane polarised.

    Option a – 57°

    Light is incident on a surface at 60°, and the reflected and refracted rays are perpendicular. The refractive index of the surface is

    a) 1.5

    b) 2

    c) 1.732

    d) None of these

    Explanation:
    This question again involves the concept of Brewster’s law and the relationship between the angle of incidence and the refractive index of a medium.

    When light strikes the boundary between two media, part of it reflects while another part refracts into the second medium. Under a specific condition known as Brewster’s angle, the reflected light becomes completely plane polarised. At this angle, the reflected and refracted rays are perpendicular to each other.

    Brewster’s law states that the tangent of the polarising angle equals the refractive index of the medium relative to the first medium. Therefore, when the angle of incidence satisfies this condition, it represents the Brewster angle for that surface.

    In the problem, the angle of incidence is given and the reflected and refracted rays are stated to be perpendicular. This confirms that the incident angle corresponds to Brewster’s angle. Using Brewster’s law, the refractive index of the surface can be determined by taking the tangent of the given angle.

    This principle is often used experimentally to determine the refractive index of transparent materials.

    In summary, when the reflected and refracted rays are perpendicular, the angle of incidence equals Brewster’s angle and the refractive index can be obtained using Brewster’s law.

    Option c – 1.732

    When unpolarised light strikes a glass plate (refractive index 1.54) at Brewster’s angle, the polarisation angle is

    a) 82°

    b) 57°

    c) 30°

    d) 60°

    Explanation:
    This question asks about the polarising angle associated with a surface of known refractive index. The polarising angle is the angle of incidence at which reflected light becomes completely plane polarised.

    When unpolarised light falls on a transparent medium, the reflected beam normally contains vibrations in many directions. However, at Brewster’s angle, only vibrations perpendicular to the plane of incidence remain in the reflected light, making it fully plane polarised.

    Brewster’s law provides the relationship between the polarising angle and the refractive index of the medium. According to this law, the tangent of the polarising angle equals the refractive index of the second medium relative to the first medium.

    To determine the polarising angle, the refractive index of the glass surface is substituted into Brewster’s law. The resulting value gives the angle at which the reflected light is completely polarised.

    This phenomenon explains why glare from surfaces such as water or glass is often partially polarised, allowing polarising filters in sunglasses or cameras to reduce the glare effectively.

    In summary, Brewster’s law links the refractive index of a material with the angle of incidence required to produce completely plane polarised reflected light.

    Option b – 57°

    Laser light is called coherent because it consists of

    a) Uncoordinated wavelengths

    b) Coordinated waves of identical wavelength

    c) Mixed wavelengths

    d) None of these

    Explanation:
    This question deals with the concept of coherence, which is an important property of laser light.

    Coherence refers to the relationship between waves that maintain a constant phase difference over time and space. When waves are coherent, their peaks and troughs occur in a consistent and predictable pattern. This property allows them to produce stable and well-defined interference patterns.

    Laser light differs from ordinary light sources because the emitted waves are highly ordered. In typical sources such as bulbs or flames, atoms emit light independently, producing waves with random phases and slightly different wavelengths. As a result, the light is incoherent.

    In a laser, however, the process of stimulated emission ensures that emitted photons have the same wavelength, direction, and phase relationship. This makes the light beam extremely coherent. Because of this coherence, laser beams can travel long distances without spreading significantly and can produce very sharp interference and Diffraction patterns.

    Applications of coherent laser light include holography, optical Communication, precision measurement, and medical procedures.

    In summary, laser light is called coherent because its waves maintain a constant phase relationship and possess nearly identical wavelengths and directions.

    Option b – Coordinated waves of identical wavelength

    The colourful fringes on floating oil on water are due to light interference. The typical thickness of the oil layer is

    a) 10 cm

    b) 100 cm

    c) 10,000 Å

    d) 100 Å

    Explanation:
    This question relates to the phenomenon of thin film interference, which is responsible for the colourful patterns often seen in oil films on water surfaces.

    When light falls on a thin transparent film such as oil, part of the light reflects from the upper surface of the film while another portion enters the film and reflects from the lower surface. These two reflected waves then combine and interfere with each other.

    The nature of the interference depends on the difference in the distances travelled by the two waves as well as the phase changes that occur upon reflection. Because the thickness of the film is extremely small, the path difference between the two waves becomes comparable to the wavelength of visible light.

    As a result, certain wavelengths interfere constructively while others interfere destructively. This selective reinforcement and cancellation of wavelengths produces the colourful appearance observed in oil films.

    The typical thickness of such films is on the order of the wavelength of visible light, which lies in the range of a few thousand angstroms or a few hundred nanometres.

    In summary, the colourful patterns arise from interference between light waves reflected from the top and bottom surfaces of a very thin oil film.

    Option c – 10,000 Å

    A sugar solution with a specific rotation of 66° has a concentration of 15 grams per 100 cc in a tube 15 cm long. The optical rotation observed is

    a) 27.8°

    b) 10.8°

    c) 14.8°

    d) 14.8°

    Explanation:
    This question concerns the measurement of optical rotation produced by an optically active substance such as a sugar solution.

    Optically active substances rotate the plane of polarised light when it passes through them. The magnitude of this rotation depends on three main factors: the specific rotation of the substance, the concentration of the solution, and the length of the tube containing the solution.

    Specific rotation is a characteristic property of the substance and indicates how strongly it rotates the plane of polarisation under standard conditions. The observed optical rotation increases when the concentration of the solution increases because more molecules interact with the light. Similarly, a longer path length means the light travels through more of the solution, producing greater rotation.

    The relationship between these quantities is expressed through a formula linking specific rotation, concentration, and path length. By substituting the known values into this relation, the total optical rotation of the light beam can be determined.

    Polarimeters commonly use this principle to determine the concentration of sugar in solutions, such as in Food and beverage industries.

    In summary, optical rotation in a solution depends directly on its specific rotation, concentration, and the length of the path through which the polarized light travels.

    Option d – 14.8°

    When a quartz wave plate is placed in a beam of linearly polarised light vibrating at 45°, the outgoing light is

    a) Vertical

    b) Vertically polarised

    c) Right circularly polarised

    d) Linearly polarised

    Explanation:
    This question examines how a wave plate affects the polarization state of light.

    A wave plate is a birefringent optical device made from materials such as quartz. When polarized light enters the wave plate, it splits into two components vibrating along two perpendicular axes known as the ordinary and extraordinary axes. These components travel through the material at different speeds because the refractive indices along these axes differ.

    As the two components propagate through the plate, a phase difference develops between them. The magnitude of this phase difference depends on the thickness of the plate and the wavelength of the light. When the incoming polarized light is oriented at a certain angle relative to the optical axes, both components are produced with equal amplitudes.

    Under specific conditions, the phase difference between these components changes the nature of the polarization of the emerging light. Depending on the phase shift produced, the output light may become circularly polarized, elliptically polarized, or remain linearly polarized.

    Wave plates are widely used in optical experiments to manipulate and control the polarization state of light.

    In summary, a wave plate modifies the relative phase of orthogonal components of polarized light, thereby changing its polarization state depending on the orientation and phase difference introduced.

    Option d – Linearly polarised

    The displacement of Raman lines from the incident light depends on the

    a) Scattering substance

    b) Intensity of incident radiation

    c) Spectrograph dispersion

    d) None of these

    Explanation:
    This question relates to Raman scattering, an important phenomenon in spectroscopy used to study Molecular structure.

    When monochromatic light interacts with molecules in a substance, most of the scattered light has the same frequency as the incident light. This is known as Rayleigh scattering. However, a small fraction of the scattered light experiences a change in frequency due to energy exchange with the vibrating molecules of the substance.

    This shifted light is known as Raman scattered light. The amount by which the frequency changes is called the Raman shift. This shift occurs because the incident photons either gain or lose energy while interacting with Molecular vibrations or rotational motions.

    The magnitude of the Raman shift depends on the energy levels associated with Molecular vibrations and rotations in the scattering material. Different substances have different Molecular structures and vibrational energy levels, so each produces a unique Raman Spectrum.

    Because of this property, Raman spectroscopy is widely used to identify substances and study Molecular structures in Chemistry, Physics, and materials science.

    In summary, the shift in Raman spectral lines depends on the Molecular vibrational energy levels of the scattering material rather than on the intensity of the incident light.

    Option a – Scattering substance

    To fully polarise reflected light on calm water, the incident light must strike at an angle of

    a) 45°

    b) 53°

    c) 0°

    d) 90°

    Explanation:
    When light falls on the boundary between two media such as air and water, part of the light is reflected and part is refracted. Under certain conditions, the reflected light becomes completely polarised. This phenomenon is associated with a specific angle of incidence known as Brewster’s angle.

    At this special angle, the reflected and refracted rays are perpendicular to each other. Because of this geometry, the component of the Electric Field vibrating parallel to the plane of incidence is not reflected. As a result, the reflected light contains vibrations only in one plane, producing plane-polarised light.

    The concept is important in Optics and explains many natural observations, such as glare from water surfaces or glass. Polarising sunglasses are designed to reduce such glare by blocking light waves vibrating in particular directions. When sunlight strikes calm water at the correct angle, the reflected light is strongly polarised, which is why glare becomes intense.

    A useful way to imagine this is sunlight hitting a smooth lake surface. At certain viewing angles, the reflected glare appears very bright because most of the reflected waves vibrate in the same direction. Polarising filters can block this component, improving visibility for activities like fishing or photography.

    Thus, the condition for complete polarisation depends on a specific incidence angle determined by the refractive indices of the two media.

    Option b – 53°

    In unpolarised light, the electric and magnetic field Vectors are arranged

    a) Perpendicularly

    b) Parallel

    c) Opposite

    d) None of these

    Explanation:
    Light is an electromagnetic wave consisting of oscillating electric and magnetic fields that travel through space. These two fields are mutually perpendicular and also perpendicular to the direction in which the wave propagates.

    In many natural light sources such as the Sun or ordinary lamps, the emitted light contains waves whose Electric Field vibrations occur in many different planes. Because of this randomness, the orientation of the Electric Field changes continuously as the wave travels. This type of radiation is referred to as unpolarised light.

    The magnetic field component is always perpendicular to the Electric Field, but its direction also changes along with the Electric Field orientation. The overall result is a beam of light whose electromagnetic vibrations occur in all possible planes perpendicular to the direction of propagation.

    To visualize this, imagine a rope being shaken randomly in different sideways directions while still moving forward. Instead of vibrating in one fixed plane, the rope’s motion keeps changing orientation. A similar effect occurs in unpolarised light where the Electric Field does not remain confined to a single plane.

    Many optical devices such as polarizers or certain crystals can restrict these vibrations to a single plane, converting unpolarised light into polarised light. Understanding this arrangement of electromagnetic fields is fundamental in Optics and wave theory.

    Option a – Perpendicularly

    Calcite crystals are classified as

    a) Opaque crystals

    b) Biaxial crystals

    c) Uniaxial crystals

    d) Transparent glass

    Explanation:
    Certain crystalline materials interact with light in unusual ways because their internal Atomic Structure is not identical in every direction. In such materials, the speed at which light travels through the crystal depends on the direction of propagation inside the crystal lattice. This directional dependence produces interesting optical effects.

    When light enters these crystals, it can split into two separate rays that travel at different velocities and along different paths. This phenomenon is known as double refraction or birefringence. One of the rays follows the ordinary laws of refraction, while the other behaves differently due to the anisotropic nature of the crystal.

    The cause of this effect lies in the arrangement of atoms or molecules inside the crystal. In some crystals, the refractive index is the same in all directions, but in others it varies with direction. When the refractive index changes with orientation, the crystal exhibits optical anisotropy and produces two distinct refracted rays.

    A familiar demonstration uses a transparent crystal placed over printed text. Instead of seeing a single image of the letters, two slightly separated images appear. This happens because each refracted ray forms its own image after passing through the crystal.

    Thus, the optical classification of crystals depends on how light propagates through their internal structure and whether the refractive index changes with direction.

    Option c – Uniaxial crystals

    If a liquid scatters light with an intensity of 6 units at 500 nm, what will the scattering intensity be at 400 nm?

    a) 12

    b) 3

    c) 6

    d) 8

    Explanation:
    Light scattering in very small particles or molecules often follows Rayleigh’s scattering law. According to this principle, the intensity of scattered light is inversely proportional to the fourth power of the wavelength of the incident radiation. This means that shorter wavelengths scatter much more strongly than longer wavelengths.

    Mathematically, the relation can be written as intensity being proportional to 1/λ4. Because the wavelength appears to the fourth power in the denominator, even a small decrease in wavelength can cause a significant increase in scattering intensity. This is why blue light, which has a shorter wavelength than red light, is scattered more effectively in Earth’s Atmosphere.

    To determine the change in intensity when the wavelength changes, the ratio of the wavelengths must be considered. By comparing the wavelengths involved and applying the inverse fourth-power relationship, the new intensity can be calculated relative to the original value.

    A common natural example is the blue color of the daytime sky. Sunlight contains many wavelengths, but the shorter blue wavelengths are scattered far more strongly by air molecules than the longer red wavelengths.

    Therefore, when the wavelength decreases from one value to another, the scattering intensity changes significantly according to the inverse fourth-power dependence described by Rayleigh scattering.

    Option b – 3

    If the Brewster angle for a material is 60°, what is the angle of refraction when light hits at that angle?

    a) 60°

    b) 45°

    c) 30°

    d) 32°

    Explanation:
    When light strikes the surface separating two transparent media, part of the light is reflected and part is refracted into the second medium. Under a specific angle of incidence known as Brewster’s angle, the reflected light becomes completely plane polarised.

    At this particular angle, an important geometric condition occurs: the reflected and refracted rays are perpendicular to each other. This condition arises from the relationship between the refractive indices of the two media and the polarization behavior of electromagnetic waves.

    The physical reason behind this effect involves the Oscillation direction of the electric field in the incident light. At Brewster’s angle, the component of the electric field that vibrates parallel to the plane of incidence cannot be reflected. As a result, the reflected beam contains vibrations in only one direction, producing plane-polarised light.

    This phenomenon explains the intense glare seen when sunlight reflects from surfaces such as water, glass, or polished roads. Polarizing sunglasses reduce this glare by filtering out light vibrating in certain directions.

    In optical analysis, the relationship between the angles involved can be derived from geometrical considerations of reflection and refraction at the interface between two media.

    Option c – 30°

    A quartz wave plate of minimum thickness t converts unpolarised light of 480 nm to circularly polarised light. For light of 600 nm, the required thickness is

    a) 0.56 t

    b) 1.25 t

    c) 1.54 t

    d) 1.5 t

    Explanation:
    Wave plates are optical devices made from birefringent materials such as quartz. These materials split incoming light into two components that travel at different speeds inside the crystal. Because of this difference in velocity, a phase difference develops between the two components as they pass through the plate.

    The amount of phase difference introduced depends on the thickness of the plate, the wavelength of the incident light, and the difference in refractive indices for the two polarization directions. For certain thicknesses, the phase difference between the two components becomes a specific fraction of a wavelength, allowing the device to transform the polarization state of light.

    When the phase difference corresponds to one-quarter of a wavelength, the plate is called a quarter-wave plate. Such a plate can convert plane-polarised light into circularly polarised light under suitable conditions. The required thickness of the plate depends directly on the wavelength used.

    If the wavelength of the incident light changes, the phase difference produced by the same plate also changes. To maintain the required polarization effect, the thickness must be adjusted in proportion to the wavelength.

    Thus, determining the new thickness involves comparing the wavelengths and applying the proportional relationship between wavelength and the optical path difference produced by the crystal.

    Option b – 1.25 t

    In light scattering, what is the ratio of scattered light intensity at 300 nm to that at 600 nm?

    a) 4

    b) 2

    c) 8

    d) 16

    Explanation:
    The scattering of light by very small particles or molecules is commonly explained using Rayleigh’s scattering law. According to this law, the intensity of scattered light depends strongly on the wavelength of the incident radiation and varies inversely with the fourth power of the wavelength.

    This relationship can be written as intensity being proportional to 1/λ4. Because of the fourth-power dependence, shorter wavelengths scatter much more strongly than longer wavelengths. Even a twofold change in wavelength can produce a very large difference in scattering intensity.

    To determine how the intensities compare for two different wavelengths, the ratio of their wavelengths must be considered and then raised to the fourth power according to the Rayleigh relation. This calculation gives the relative scattering strength between the two wavelengths.

    A well-known natural consequence of this effect is the blue color of the daytime sky. Sunlight contains many wavelengths, but the shorter blue wavelengths are scattered more efficiently by atmospheric molecules than the longer red wavelengths.

    Therefore, when comparing scattering at two wavelengths, the relative intensities can be determined by applying the inverse fourth-power dependence described in Rayleigh scattering theory.

    Option d – 16

    Which of the following is an example of a Boson?

    a) Electron

    b) Neutron

    c) Proton

    d) Photon

    Explanation:
    In quantum Physics, elementary particles are classified into two major categories based on their intrinsic angular momentum, or spin. These categories are known as fermions and bosons. The classification determines how groups of identical particles behave when they occupy the same quantum system.

    Bosons are particles that possess integer values of spin. Because of this property, they follow a statistical distribution known as Bose–Einstein statistics. One of the key features of bosons is that many identical particles can occupy the same quantum state simultaneously.

    This behavior leads to several remarkable physical phenomena. For instance, certain particles can form collective quantum states where a large number of identical particles behave as a single coherent entity. Examples of such phenomena include laser light and Bose–Einstein condensates observed at extremely low temperatures.

    In contrast, fermions have half-integer spin and obey the Pauli exclusion principle, which prevents two identical fermions from occupying the same quantum state at the same time.

    Understanding the distinction between these particle types is fundamental to modern quantum theory and explains the behavior of Matter and radiation at microscopic scales.

    Option d – Photon

    The electron is classified as which type of particle? ( CSIR NET Physics Previous Year Question Papers mcq )

    a) Boson

    b) Fermion

    c) Both

    d) None

    Explanation:
    Elementary particles are often classified according to their intrinsic spin and the statistical rules they follow. One important group of particles has half-integer values of spin, such as 1/2, 3/2, and so on. These particles follow a statistical behavior described by Fermi–Dirac statistics.

    Particles in this category obey a fundamental rule known as the Pauli exclusion principle. This principle states that no two identical particles of this type can occupy the same quantum state simultaneously within a quantum system. This rule plays a crucial role in determining the structure of atoms and the arrangement of electrons in atomic orbitals.

    Because of this restriction, electrons fill available energy levels in atoms one by one, following specific rules of quantum mechanics. The resulting arrangement explains the Periodic Table and the chemical properties of elements.

    In contrast, particles with integer spin values follow a different statistical rule that allows multiple particles to share the same quantum state. The distinction between these two particle categories forms one of the fundamental classifications in particle Physics.

    Thus, the electron belongs to the group of particles characterized by half-integer spin and exclusion-based statistical behavior.

    Option b – Fermion

    Which particles obey Pauli’s exclusion principle? ( CSIR NET Physics Previous Year Question Papers mcq )

    a) Fermions

    b) Bosons

    c) Both

    d) None

    Explanation:
    The Pauli exclusion principle is one of the most important rules in quantum mechanics. It states that no two identical particles with certain quantum properties can occupy the same quantum state simultaneously within a system. This rule strongly influences the structure of atoms and the behavior of Matter.

    Particles that obey this principle possess half-integer spin values, such as 1/2 or 3/2. Because of their spin characteristics, these particles follow Fermi–Dirac statistics. Their wave functions are antisymmetric, meaning that exchanging two identical particles results in a sign change in the overall quantum state.

    This property prevents multiple identical particles from occupying the same SET of quantum numbers in atoms. As a consequence, electrons fill atomic orbitals in a specific order, giving rise to well-defined electronic configurations and the structure of the Periodic Table.

    In contrast, particles with integer spin follow Bose–Einstein statistics and are not restricted by the exclusion principle. Many identical particles of that type can occupy the same quantum state simultaneously.

    The distinction between these two statistical behaviors forms a fundamental basis for understanding Atomic Structure, Solid-state Physics, and many phenomena in modern quantum theory.

    Option a – Fermions

    The thermoelectric power of iron is 1734 µV at 0°C and 1247 µV at 100°C. Copper has 136 µV at 0°C and 231 µV at 100°C. Calculate the emf of the iron-copper pair between 0°C and 100°C.

    a) 17.4 V

    b) 130.7 mV

    c) 130.76 V

    d) None of these

    Explanation:
    The thermoelectric effect occurs when two different conductors are joined to form a closed circuit and their junctions are kept at different temperatures. Under these conditions, an electromotive force (emf) is generated due to the temperature difference between the junctions. This phenomenon is known as the Seebeck effect.

    Each metal possesses a characteristic thermoelectric power that varies with temperature. When two Metals are combined to form a thermocouple, the NET emf produced depends on the difference between their thermoelectric powers at the specified temperatures.

    To determine the emf generated between two junction temperatures, the thermoelectric power values for each metal at those temperatures must be considered. The difference between the values for the two Metals determines the effective thermoelectric contribution of the pair.

    By comparing the thermoelectric powers of the two Metals at both temperatures and applying the appropriate relationship, the emf produced by the thermocouple can be calculated.

    Thermocouples based on this principle are widely used for temperature measurements, especially in industrial processes and scientific instruments where reliable temperature sensing is required over a wide range.

    Option b – 130.7 mV

    Which law states that the total energy emitted per unit area per second by a perfect black body is proportional to the fourth power of its absolute temperature? ( CSIR NET Physics Previous Year Question Papers mcq )

    a) Wien’s Law

    b) Newton’s Law

    c) Stefan’s Law

    d) Stefan-Boltzmann Law

    Explanation:
    A perfect black body is an idealized physical object that absorbs all incident radiation regardless of wavelength or direction. When such an object is heated, it emits electromagnetic radiation whose intensity and spectral distribution depend solely on its temperature.

    One of the key principles describing this radiation relates the total energy emitted by the black body to its absolute temperature. According to this principle, the radiant energy emitted per unit surface area per unit time increases rapidly as the temperature rises.

    Mathematically, the relationship shows that the emitted energy varies with the fourth power of the absolute temperature, often written as proportional to T4. Because of this fourth-power dependence, even a moderate increase in temperature leads to a large increase in emitted radiation.

    This law helps explain many physical phenomena, including the glowing of heated Metals, the radiation from stars, and the thermal emission from hot surfaces. Astronomers frequently use this relationship to estimate the energy output of stars based on their surface temperatures.

    Thus, the connection between radiated energy and the fourth power of temperature forms a fundamental principle in thermal radiation theory.

    Option c – Stefan’s Law

    For the thermoelectric power of iron vs lead given as (1734 – 4.87T) and for copper as (136 + 0.95T), calculate the emf of an iron-copper junction between 20°C and 100°C.

    a) 909.2 mV

    b) 900 V

    c) 1000 mV

    d) 90 V

    Explanation:
    Thermoelectric power represents the emf generated per degree change in temperature for a given metal. In a thermocouple made from two different Metals, the total emf produced between two junction temperatures depends on the difference in thermoelectric power of the Metals across that temperature range.

    The thermoelectric power of each metal can vary with temperature and is often expressed as a linear function of temperature. When two such functions are given, the effective thermoelectric power of the thermocouple is obtained by subtracting one expression from the other.

    Once the effective thermoelectric power of the pair is known, the emf generated between two temperatures can be calculated by integrating the thermoelectric power over the temperature interval. This approach accounts for the variation of thermoelectric properties with temperature rather than assuming a constant value.

    Thermocouples based on this principle are widely used for accurate temperature measurements. They are especially useful in industrial furnaces, laboratories, and engineering systems where temperatures can vary significantly.

    Thus, by combining the temperature-dependent expressions of thermoelectric power and evaluating the temperature interval involved, the emf produced by the iron–copper junction can be determined.

    Option a – 909.2 mV

    Which Metals are used in thermocouples for high-temperature measurements? ( CSIR NET Physics Previous Year Question Papers mcq )

    a) Platinum

    b) Molybdenum

    c) Iron

    d) Both a and b

    Explanation:
    Thermocouples are temperature-measuring devices that operate on the thermoelectric effect. They consist of two dissimilar metal wires joined together at one end, forming a junction. When the junction is exposed to a temperature different from the other ends, an electromotive force is produced.

    For high-temperature applications, the Metals used must satisfy several important requirements. They should have high melting points, good mechanical strength at elevated temperatures, and stable thermoelectric properties over a wide temperature range. Resistance to oxidation and chemical reactions is also important for reliable long-term operation.

    Certain combinations of Metals or metal alloys are specially designed to meet these conditions. These materials maintain consistent thermoelectric characteristics even at very high temperatures, making them suitable for use in furnaces, power plants, and industrial processing equipment.

    For example, thermocouples used in metallurgical industries or high-temperature laboratories must withstand extremely hot environments without degrading or producing unstable readings. The choice of Metals therefore depends on both their physical durability and their thermoelectric behavior.

    Selecting appropriate thermocouple materials ensures accurate temperature measurement in demanding high-temperature conditions.

    Option d – Both a and b

    In a double-slit setup with white light, when the screen is moved 0.05 m away, what happens to the central white fringe?

    a) It disappears

    b) It becomes coloured

    c) It remains stationary

    d) It shifts from its original position

    Explanation:
    Young’s double slit experiment demonstrates the interference of light waves originating from two coherent sources. When white light is used, the interference pattern consists of a bright central region surrounded by colored fringes due to the presence of many wavelengths.

    The central fringe appears white because the path difference between the two waves at that location is zero. As a result, all wavelengths present in the white light undergo constructive interference simultaneously, producing a bright white region.

    The spacing between interference fringes depends on the wavelength of light, the distance between the slits, and the distance from the slits to the observation screen. When the screen is moved farther from the slits, the geometry of the setup changes and the fringe spacing increases.

    Because of this change in spacing, the interference pattern spreads out across the screen. However, the condition for the central fringe remains unchanged since it corresponds to zero path difference between the waves from the two slits.

    Therefore, changing the screen distance mainly affects the spacing and visibility of the fringes while the central region continues to form due to the constructive overlap of all wavelengths.

    Option c – It remains stationary

    What does a fringe represent in interference patterns?

    a) Constant amplitude

    b) Constant phase

    c) Constant wavelength

    d) None of these

    Explanation:
    Interference occurs when two or more coherent waves overlap in space. As these waves combine, their amplitudes add together according to the principle of superposition. Depending on the phase relationship between the waves, the result may be reinforcement or cancellation.

    In optical interference experiments such as Young’s double slit setup, this superposition produces alternating bright and dark regions on a screen. These regions appear as parallel bands or patterns that repeat at regular intervals.

    Each bright band corresponds to a position where the waves arrive in phase and reinforce one another. Conversely, each dark band represents a location where the waves arrive out of phase and cancel each other partially or completely.

    The distance between successive fringes depends on several factors, including the wavelength of light, the separation between coherent sources, and the distance from the sources to the observation screen.

    A simple analogy is the pattern formed when two sets of water waves intersect on the surface of a pond. Some areas appear calm while others show larger wave motion because of constructive and destructive combinations.

    Thus, fringes represent the spatial distribution of intensity created by the superposition of coherent waves.

    Option b – Constant phase

    In Fresnel’s biprism experiment, virtual coherent sources are produced through which process? ( CSIR NET Physics Previous Year Question Papers mcq )

    a) Reflection

    b) Refraction

    c) Dispersion

    d) Interference

    Explanation:
    The Fresnel biprism experiment is designed to demonstrate the interference of light using a single real source. Since interference requires two coherent sources with a constant phase relationship, the optical system must generate two equivalent sources from the original one.

    A Fresnel biprism consists of two thin prisms joined together at their Bases. When light from a narrow slit passes through the biprism, the rays are refracted slightly in opposite directions by the two prism halves.

    Because of this refraction, the light emerging from the biprism appears to originate from two separate virtual sources located behind the prism. These sources are formed from the same original slit, which ensures that the waves emerging from them maintain a constant phase relationship.

    The interference pattern observed on a screen arises when the waves from these two virtual sources overlap. The resulting pattern consists of alternating bright and dark fringes that demonstrate the wave Nature of Light.

    This arrangement allows interference to be observed without the need for two physically separate light sources, making the experiment a classic demonstration in wave Optics.

    Option b – Refraction

    The refracting angle of Fresnel’s biprism is approximately

    a) 1/2°

    b) 180°

    c) 179°

    d) 90°

    Explanation:
    A Fresnel biprism is an optical device formed by joining two very thin prisms together at their Bases. The purpose of this arrangement is to split light from a single narrow slit into two coherent beams that can interfere with each other.

    In order to produce clear interference fringes, the angular deviation caused by the prism must be very small. If the prism angle were large, the two beams would diverge too much and would not overlap effectively on the screen.

    Therefore, the biprism is designed with an extremely small refracting angle. This small angle ensures that the two refracted beams remain close enough to overlap and create an interference pattern within a reasonable distance from the prism.

    Because the deviation is slight, the virtual sources produced by the biprism appear only a small distance apart. This close separation is essential for producing observable interference fringes with measurable spacing.

    Thus, the refracting angle of a Fresnel biprism is deliberately made very small so that the overlapping beams form a stable and visible interference pattern.

    Option a – 1/2°

    Light waves have properties similar to which of the following? ( CSIR NET Physics Previous Year Question Papers mcq )

    a) Alpha rays

    b) Cathode rays

    c) Gamma rays

    d) Cosmic rays

    Explanation:
    Light is an electromagnetic wave that exhibits several behaviors characteristic of wave motion. These include reflection, refraction, Diffraction, and interference. Such phenomena arise because light propagates as oscillating electric and magnetic fields through space.

    The wave Nature of Light became widely accepted after experiments demonstrated patterns that could only be explained through wave superposition. For instance, interference patterns in double-slit experiments and Diffraction patterns around obstacles clearly show behavior typical of waves.

    In these situations, light waves interact with boundaries and openings in ways that resemble other types of wave motion observed in nature. The ability to spread out after passing through a narrow opening or to combine with other waves to form patterns is a strong indication of its wave-like character.

    A common comparison can be made with waves traveling across the surface of water. When two sets of water waves meet, they produce alternating regions of larger and smaller wave heights due to constructive and destructive superposition.

    Thus, the fundamental properties of light reveal that it behaves similarly to other wave phenomena observed in physics.

    Option c – Gamma rays

    Two coherent monochromatic light beams with intensities I and 4I are superimposed. The maximum and minimum resultant intensities are

    a) 5I and I

    b) 5I and 3I

    c) 9I and I

    d) 9I and 3I

    Explanation:
    When two coherent light waves overlap, their amplitudes combine according to the principle of superposition. The resulting intensity at any point depends on the phase difference between the waves at that location.

    Intensity in a wave is proportional to the square of its amplitude. If two waves have different intensities, their amplitudes are related to the square root of those intensities. When the waves arrive in phase, their amplitudes add together, producing the largest possible intensity known as constructive interference.

    When the waves arrive exactly out of phase, their amplitudes partially cancel. This produces the smallest possible intensity, known as destructive interference. The difference between maximum and minimum intensities depends on the amplitude ratio of the two waves.

    By converting the given intensities into amplitudes and applying the interference relations, the resulting maximum and minimum intensities can be determined.

    This principle explains the bright and dark fringes observed in interference experiments, where varying phase differences across the screen create a repeating pattern of intensity variations.

    Option c – 9I and I

    In interference, fringe width is ( CSIR Net Physics Previous Year Question Papers mcq )

    a) Directly proportional to slit width

    b) Inversely proportional to slit separation

    c) Inversely proportional to screen distance

    d) All of these

    Explanation:
    In interference experiments such as Young’s double slit setup, alternating bright and dark bands appear on the observation screen. These bands form because of constructive and destructive interference between waves coming from two coherent sources.

    The distance between two successive bright fringes or two successive dark fringes is known as the fringe width. This quantity depends on several experimental parameters, including the wavelength of the light used, the separation between the two slits, and the distance between the slits and the screen.

    Mathematically, the fringe width is proportional to the wavelength and the distance to the screen, and inversely proportional to the separation between the slits. This means that increasing the wavelength or moving the screen farther away increases the spacing between fringes.

    Conversely, if the slit separation becomes larger, the fringes become more closely spaced. This relationship allows experimental measurement of wavelengths and other optical properties.

    Thus, fringe width represents the spacing between successive interference bands produced by coherent light waves.

    Option b – Inversely proportional to slit separation

    In which setup are coherent sources created using wavefront division?

    a) Fresnel’s biprism

    b) Lloyd’s mirror

    c) Young’s double slit

    d) Lloyd’s mirror only

    Explanation:
    Coherent sources are required to produce stable interference patterns. Two sources are considered coherent when they emit waves with the same frequency and maintain a constant phase difference over time.

    One method used in Optics to obtain coherent sources is called division of wavefront. In this method, light from a single source is divided into separate parts of the same wavefront. These parts then act as individual coherent sources because they originate from the same original wave.

    Optical devices such as narrow slits, biprisms, or mirrors can be arranged so that the wavefront is split into two portions. These portions propagate along slightly different paths and later overlap to produce interference patterns.

    The advantage of this technique is that it ensures the required phase relationship between the resulting sources. Since both parts come from the same initial wavefront, their coherence is naturally maintained.

    Experiments demonstrating interference using this method have played an important role in establishing the wave Nature of Light and in measuring wavelengths accurately.

    Option b – Lloyd’s mirror

    In Newton’s rings, the diameters of successive dark rings vary ( CSIR Net Physics Previous Year Question Papers mcq )

    a) Directly with natural numbers

    b) Directly with the square of natural numbers

    c) Directly with the square root of natural numbers

    d) Inversely with natural numbers

    Explanation:
    Newton’s rings are circular interference patterns produced when a convex lens is placed on a flat glass plate. A thin film of air forms between the curved surface of the lens and the plate. When monochromatic light is incident on this arrangement, reflections occur from the top and bottom surfaces of the air film.

    Because the thickness of the air film changes gradually from the point of contact outward, the path difference between the reflected rays also changes with distance from the center. This varying path difference leads to alternating constructive and destructive interference, forming bright and dark circular rings.

    For dark rings, the condition for destructive interference depends on the thickness of the air film and the wavelength of light. The mathematical relation connecting the ring diameter with the order number shows that the square of the diameter is proportional to the order of the ring.

    This relationship means the rings do not increase in diameter by equal amounts as one moves outward. Instead, the spacing between successive rings gradually changes according to the square-root dependence.

    As a result, the pattern exhibits a predictable variation in ring size that can be used experimentally to determine the wavelength of light or other optical parameters.

    Option c – Directly with the square root of natural numbers

    In an air wedge formed by two glass plates, a wire of 0.05 mm diameter separates them at a distance of 15 cm. Using monochromatic light of wavelength 6000 Å, the fringe width is

    a) 0.02 cm

    b) 0.06 cm

    c) 0.03 cm

    d) 0.09 cm

    Explanation:
    An air wedge is formed when two glass plates are placed together with one end slightly separated by a thin spacer such as a wire. This arrangement produces a wedge-shaped air film whose thickness increases gradually along the length of the plates.

    When monochromatic light falls on this wedge, interference occurs between the light waves reflected from the upper and lower surfaces of the thin air film. Because the thickness changes linearly along the wedge, the path difference between the interfering waves also changes steadily.

    This produces a series of parallel bright and dark fringes. The spacing between these fringes, called the fringe width, depends on the wavelength of light, the length of the wedge, and the thickness of the air film at the separated end.

    In such setups, the thickness difference introduced by the wire determines the wedge angle. By relating this angle to the wavelength and geometry of the system, the fringe spacing can be calculated.

    Air wedge experiments are commonly used in Optics laboratories to measure very small thicknesses or to determine the wavelength of light with high precision.

    Option d – 0.09 cm

    In Young’s double slit experiment, if the red light source is replaced with violet light, what happens?

    a) Bands become brighter

    b) Bands move closer

    c) Intensities increase

    d) None of these

    Explanation:
    Young’s double slit experiment produces an interference pattern consisting of alternating bright and dark fringes on a screen. These fringes arise from the constructive and destructive interference of light waves coming from two closely spaced slits.

    The spacing between adjacent fringes depends on the wavelength of the light used, the separation between the slits, and the distance between the slits and the observation screen. Among these factors, the wavelength plays an important role in determining how widely the fringes appear.

    Red light has a longer wavelength than violet light in the visible Spectrum. When the wavelength changes while the slit separation and screen distance remain constant, the fringe spacing changes proportionally.

    This means that shorter wavelengths produce a more closely spaced interference pattern, while longer wavelengths produce wider spacing between fringes.

    Therefore, replacing one color of light with another alters the geometry of the observed pattern because the interference condition depends directly on the wavelength of the light involved.

    Option b – Bands move closer

    In a Michelson interferometer, the angle between the mirrors is

    a) 45°

    b) 90°

    c) 180°

    d) 360°

    Explanation:
    The Michelson interferometer is an optical instrument used to study interference and measure extremely small distances or wavelength changes. It works by splitting a beam of light into two parts that travel along different paths before recombining to produce interference.

    Inside the interferometer, a beam splitter divides the incoming light into two perpendicular beams. Each beam travels toward a mirror placed in its path. After reflection, the beams return to the beam splitter and recombine, producing an interference pattern.

    For the device to function properly, the mirrors must be arranged so that the returning beams overlap accurately. The geometry of the setup is designed so that the optical paths are carefully controlled and aligned.

    In many applications, the mirrors are positioned in a way that maintains the required symmetry between the two light paths. Small adjustments of the mirrors can change the interference pattern, allowing precise measurement of wavelength differences or refractive index variations.

    This careful alignment of mirrors is essential for producing stable and observable interference fringes in the Michelson interferometer.

    Option b – 90°

    Newton’s rings observed in transmitted light appear as ( CSIR Net Physics Previous Year Question Papers mcq )

    a) Ellipses

    b) Similar to reflected light rings

    c) Concentric rings

    d) Hyperbolas

    Explanation:
    Newton’s rings are interference patterns formed due to the thin air film created between a convex lens and a flat glass plate. When monochromatic light is incident on the system, interference occurs between light waves reflected from the upper and lower surfaces of the air film.

    When the pattern is observed in reflected light, alternating bright and dark circular rings appear due to constructive and destructive interference of the reflected rays. However, the appearance of the pattern changes when it is viewed in transmitted light.

    In transmitted light, the interference occurs between waves that have passed through the thin air film rather than those reflected from its surfaces. Because the phase relationships differ between reflected and transmitted waves, the intensity distribution of the rings changes.

    As a result, the positions that appear dark in the reflected pattern correspond to bright regions in the transmitted pattern, and vice versa. This complementary behavior arises from the energy distribution between reflected and transmitted light.

    Thus, the interference pattern observed in transmitted light is related to the reflected pattern but shows a reversed intensity arrangement.

    Option c – Concentric rings

    An air wedge with an angle of 0.02 mm radian is illuminated by monochromatic light. If the fringe width is 0.015 mm, the wavelength used is

    a) 6000 Å

    b) 4000 Å

    c) 5000 Å

    d) 3000 Å

    Explanation:
    In an air wedge experiment, two glass plates form a very small angle between them, producing a thin film of air whose thickness increases gradually along the wedge. When monochromatic light is incident on this arrangement, interference occurs between the light waves reflected from the upper and lower surfaces of the air film.

    Because the thickness changes linearly with distance along the wedge, the path difference between the interfering rays also changes steadily. This produces a pattern of parallel bright and dark fringes whose spacing depends on the wavelength of the light and the wedge angle.

    The fringe width in an air wedge is determined by a relationship that connects the wavelength of light with the wedge angle. When the fringe width and wedge angle are known, the wavelength can be calculated using this interference relation.

    Such experiments are commonly performed in Optics laboratories because they allow measurement of very small wavelengths using simple geometric arrangements.

    Therefore, by applying the fringe-width relation for an air wedge, the wavelength corresponding to the observed interference pattern can be determined.

    Option a – 6000 Å

    The central spot in Newton’s rings seen by reflected light appears ( CSIR Net Physics Previous Year Question Papers mcq )

    a) Black

    b) White

    c) Bright

    d) Coloured

    Explanation:
    In the Newton’s rings experiment, a convex lens placed on a flat glass plate creates a thin air film whose thickness increases gradually from the point of contact outward. When monochromatic light falls on this system, interference occurs between the light waves reflected from the two surfaces of the air film.

    At the point where the lens touches the glass plate, the thickness of the air film is essentially zero. Because of this, the path difference between the reflected waves is determined mainly by the phase change that occurs upon reflection.

    When light reflects from a boundary between a rarer and a denser medium, a phase shift of half a wavelength may occur. This phase change affects the interference condition at the center of the pattern.

    As a result, the interference condition at the point of contact produces a particular intensity behavior that differs from nearby regions where the air film has finite thickness.

    This explains why the central region of the Newton’s rings pattern shows a distinctive intensity compared with the surrounding rings.

    Option a – Black

    Newton’s rings observed through transmitted light appear as

    a) Elliptical shapes

    b) Similar rings as seen in reflected light

    c) Circular rings

    d) Hyperbolic patterns

    Explanation:
    The Newton’s rings pattern arises from interference in a thin air film formed between a convex lens and a flat glass plate. When monochromatic light falls on the system, some light is reflected while some passes through the air film and emerges from the other side.

    In reflected light, interference occurs between rays reflected from the upper and lower surfaces of the air film. This produces alternating bright and dark rings depending on the local thickness of the film.

    However, when the pattern is viewed in transmitted light, the interfering waves are those that have passed through the thin film rather than those reflected. Because the energy distribution between reflection and transmission differs, the resulting intensity pattern changes.

    The positions that appear dark in reflected light correspond to regions where the transmitted light intensity is relatively high. Similarly, regions that are bright in reflection correspond to reduced transmitted intensity.

    Therefore, the interference pattern observed in transmitted light forms a complementary arrangement relative to the pattern seen in reflected light.

    Option c – Circular rings

    The mirrors in a Michelson interferometer are positioned at an angle of ( CSIR Net Physics Previous Year Question Papers MCQ )

    a) 45°

    b) 90°

    c) 180°

    d) 360°

    Explanation:
    The Michelson interferometer is designed to split a single beam of light into two separate beams that travel along different paths. These beams are later recombined to produce interference patterns that can reveal extremely small differences in optical path length.

    The instrument uses a partially reflecting beam splitter placed at a specific orientation to divide the incoming beam. After splitting, the two beams travel toward mirrors positioned along perpendicular directions.

    Each mirror reflects the beam back toward the beam splitter, where the two returning beams overlap and interfere. The interference pattern formed depends on the difference in optical path lengths between the two arms of the interferometer.

    Careful alignment of the mirrors is necessary to ensure that the beams recombine correctly and produce clear interference fringes. Small adjustments in mirror orientation can change the pattern, allowing precise measurement of wavelength or displacement.

    The geometry of the mirrors relative to each other is therefore essential for maintaining the correct interference conditions in the interferometer.

    Option b – 90°

    Which of the following devices offers the highest resolving power?

    a) Prism

    b) Diffraction Grating

    c) Telescope

    d) Fabry-Perot Interferometer

    Explanation:
    Resolving power refers to the ability of an optical instrument to distinguish between two closely spaced spectral lines or details. In spectroscopy and optical analysis, high resolving power is essential for studying fine structural features of light.

    Different optical instruments achieve resolution using different physical principles. Some rely on refraction through prisms, while others use Diffraction from finely spaced lines or grooves.

    Diffraction-based devices generally provide greater resolving capability because they separate wavelengths based on interference effects from multiple coherent sources. When a large number of closely spaced lines are used, the resulting interference pattern becomes extremely sharp.

    This increased sharpness allows the instrument to distinguish between wavelengths that differ by very small amounts. Because of this property, Diffraction-based instruments are widely used in spectroscopy and precision optical measurements.

    Thus, instruments that utilize Diffraction and interference effects tend to achieve very high resolving power compared with simpler optical devices.

    Option d – Fabry-Perot Interferometer

    In which experiment are interference patterns formed due to the division of wavefronts?

    a) Young’s Double Slit Experiment

    b) Fresnel’s Biprism Experiment

    c) Lloyd’s Mirror Experiment

    d) Newton’s Rings

    Explanation:
    Interference of light occurs when two coherent waves overlap and combine according to the principle of superposition. For stable interference patterns to appear, the light sources must maintain a constant phase relationship and have the same frequency.

    One way to obtain such coherent sources is through a technique known as division of wavefront. In this method, light from a single source is divided into two or more portions of the same wavefront. Since these portions originate from the same source at the same instant, they naturally maintain coherence.

    Optical arrangements using narrow slits, biprisms, or mirrors can split the wavefront into separate parts that travel along slightly different paths. When these parts overlap again on a screen, they produce alternating bright and dark fringes due to constructive and destructive interference.

    A useful analogy is the pattern formed when ripples from two points on a water surface intersect. Where the waves reinforce each other, larger ripples appear, and where they cancel, the water surface becomes relatively calm.

    Thus, interference experiments that divide the original wavefront into separate coherent portions demonstrate the wave Nature of Light and produce observable fringe patterns.

    Option d – Newton’s Rings

    If Young’s double slit experiment is done in air and then repeated in water, the fringe spacing will

    a) Remain unchanged

    b) Decrease

    c) Increase

    d) Disappear completely

    Explanation:
    Young’s double slit experiment produces an interference pattern consisting of alternating bright and dark fringes. The spacing between these fringes, called fringe width, depends on the wavelength of light used, the distance between the slits and the screen, and the separation between the slits.

    When the experiment is performed in a medium other than air, the speed of light changes according to the refractive index of that medium. Although the frequency of light remains unchanged, the wavelength decreases in a medium with a higher refractive index.

    Because fringe width depends directly on wavelength, any change in wavelength will influence the spacing between interference fringes. When the experiment is repeated in water, the optical properties of the medium alter the effective wavelength of the light passing through it.

    This change affects the geometry of the interference pattern observed on the screen. The fringes become either closer together or farther apart depending on how the wavelength is modified by the medium.

    Therefore, the surrounding medium plays an important role in determining the spacing of the interference fringes produced in the double slit experiment.

    Option b – Decrease

    Two coherent sources have intensity ratios of 81:1. The ratio of maximum to minimum intensity in the interference pattern will be ( CSIR Net Physics Previous Year Question Papers MCQ )

    a) 8:1

    b) 9:1

    c) 10:8

    d) 25:16

    Explanation:
    In interference, the resultant intensity at any point depends on the amplitudes of the waves and the phase difference between them. The intensity of a wave is proportional to the square of its amplitude, meaning that amplitude can be determined by taking the square root of the intensity.

    When two coherent waves interfere constructively, their amplitudes add together, producing the maximum possible intensity. When they interfere destructively, the amplitudes partially cancel, resulting in the minimum intensity.

    To analyze the interference pattern when the intensities are different, the first step is converting the given intensity ratio into an amplitude ratio. Since amplitude is proportional to the square root of intensity, the amplitude relationship becomes much simpler to work with.

    Once the amplitude ratio is known, the expressions for maximum and minimum intensities can be written in terms of these amplitudes. These expressions reveal how strongly the bright and dark fringes differ in intensity.

    Such calculations help determine the contrast or visibility of interference patterns observed in optical experiments.

    Option d – 25:16

    If two coherent sources have intensities I and 4I, the ratio of maximum to minimum intensity in their interference pattern will be

    a) 9:1

    b) 4:1

    c) 16:1

    d) 1:1

    Explanation:
    When two coherent light waves overlap, the resulting intensity depends on both their amplitudes and the phase difference between them. Constructive interference occurs when the waves arrive in phase, while destructive interference occurs when their phases differ by half a wavelength.

    The relationship between amplitude and intensity is important in interference calculations. Since intensity is proportional to the square of amplitude, the amplitudes of the two waves can be obtained by taking the square root of their respective intensities.

    Once the amplitudes are known, the maximum intensity occurs when the amplitudes combine directly through constructive interference. The minimum intensity occurs when the amplitudes subtract due to destructive interference.

    Using these amplitude relations, expressions for maximum and minimum intensities can be derived. These expressions show how differences in the original intensities affect the contrast between bright and dark fringes in the interference pattern.

    Such analysis explains why some interference patterns appear very bright and clear, while others show weaker contrast between the fringes.

    Option a – 9:1

    Which scientist first demonstrated the interference of light? ( CSIR Net Physics Previous Year Question Papers MCQ )

    a) Foucault

    b) Huygens

    c) Young

    d) Newton

    Explanation:
    During the early study of Optics, scientists debated whether light behaved as a stream of particles or as a wave. Many early theories favored the particle model because it explained reflection and refraction reasonably well.

    However, certain optical phenomena such as Diffraction and interference could not be fully explained using only the particle theory. These effects suggested that light behaves more like a wave that can combine and interact with other waves.

    A landmark experiment demonstrated that when light from a single source passes through two closely spaced openings, it produces alternating bright and dark bands on a screen. These bands arise from constructive and destructive interference of light waves.

    This experiment provided strong evidence that light behaves as a wave capable of superposition. It became one of the most important demonstrations supporting the wave theory of light.

    The discovery significantly influenced the development of modern Optics and helped establish the wave nature of electromagnetic radiation.

    Option c – Young

    Which of the following can function as coherent sources?

    a) A 60W bulb and a 100W bulb

    b) Two separate 60W bulbs

    c) Two halves of a 60W bulb

    d) Two virtual sources created by Fresnel’s Biprism

    Explanation:
    Coherent sources are essential for producing stable and clearly visible interference patterns. Two sources are said to be coherent when they emit waves of the same frequency while maintaining a constant phase difference over time.

    Independent light sources, such as two separate lamps, generally do not maintain a fixed phase relationship. Because their emissions occur randomly, the interference pattern produced by them changes rapidly and becomes unobservable.

    To obtain coherent sources, optical experiments usually derive multiple beams from a single original source. By splitting a single wavefront or dividing the amplitude of the wave, two or more beams can be created that maintain a stable phase relationship.

    Devices such as double slits, biprisms, and beam splitters are commonly used to produce such coherent beams. Since these beams originate from the same source, they retain the same frequency and a fixed phase relationship.

    This principle ensures that interference fringes remain steady and visible on the observation screen.

    Option d – Two virtual sources created by Fresnel’s Biprism

    The interference pattern is a direct result of which fundamental principle?

    a) Conservation of Energy

    b) Conservation of Momentum

    c) Conservation of Angular Momentum

    d) Rectilinear Propagation of Light

    Explanation:
    Interference patterns arise when two or more waves overlap and combine at the same point in space. The resulting displacement at that point depends on how the individual waves add together.

    This behavior is explained by a fundamental rule in wave physics known as the principle of superposition. According to this principle, when multiple waves meet, the resultant displacement is equal to the algebraic sum of the displacements produced by each wave individually.

    If two waves arrive at a point with the same phase, their displacements reinforce each other, producing a region of greater intensity. When the waves arrive out of phase, their displacements partially or completely cancel, creating regions of lower intensity.

    In optical experiments, these variations in intensity appear as bright and dark fringes on a screen. The entire interference pattern therefore results from the repeated constructive and destructive combination of waves.

    This principle applies not only to light waves but also to other types of waves such as sound waves and water waves.

    Option a – Conservation of Energy

    If the amplitude ratio of two superimposed waves is 2:1, the ratio of maximum to minimum intensity will be ( CSIR Net Physics Previous Year Question Papers MCQ )

    a) 2:1

    b) 4:1

    c) 3:1

    d) 9:1

    Explanation:
    When two waves interfere, the resulting intensity pattern depends on the amplitudes of the waves and the phase difference between them. Constructive interference occurs when the waves reinforce each other, while destructive interference occurs when they partially cancel.

    Intensity in a wave is proportional to the square of its amplitude. Therefore, if the amplitudes of two waves are known, their corresponding intensities can be determined using this square relationship.

    During constructive interference, the amplitudes add together, producing the largest possible resultant amplitude. During destructive interference, the amplitudes subtract from each other, producing the smallest resultant amplitude.

    By applying these amplitude relations, expressions for maximum and minimum intensities can be derived. The ratio of these intensities depends on the relative amplitudes of the interfering waves.

    This relationship explains why some interference patterns have strong contrast between bright and dark fringes while others appear less distinct.

    Option d – 9:1

    In which interference setup is a real source combined with its virtual image to create the pattern?

    a) Young’s Double Slit Experiment

    b) Lloyd’s Mirror Experiment

    c) Fresnel’s Biprism Experiment

    d) Fresnel’s Double Mirror Experiment

    Explanation:
    In some optical interference experiments, coherent sources are produced not by dividing the wavefront directly but by creating virtual images of a real source. Optical elements such as mirrors or prisms can produce these virtual images through reflection or refraction.

    When light from a real source interacts with such optical components, it may appear to originate from another point in space. This point acts as a virtual source because the light rays seem to diverge from it even though no physical source exists there.

    If the real source and its virtual image emit waves that overlap on a screen, interference can occur because both waves originate from the same original source. Their phase relationship therefore remains constant.

    The overlapping waves produce a pattern of alternating bright and dark regions due to constructive and destructive interference.

    This method of producing coherent sources is commonly used in several classic optical experiments designed to demonstrate the wave Nature of Light.

    Option b – Lloyd’s Mirror Experiment

    When light reflects off a denser medium, the additional phase shift is ( CSIR Net Physics Previous Year Question Papers MCQ )

    a) π/2

    b) 2π

    c) 3π/2

    d) π

    Explanation:
    When a light wave encounters a boundary between two different media, part of the wave may be reflected while the rest is transmitted. The behavior of the reflected wave depends on the optical properties of the two media involved.

    If light reflects from a boundary where the second medium is optically denser than the first, the reflected wave undergoes a change in phase. This phase change occurs because the electromagnetic wave interacts with the electrons in the denser medium.

    The phase change plays an important role in thin-film interference phenomena such as Newton’s rings, soap bubbles, and oil films. In these situations, the phase shift affects the condition for constructive or destructive interference.

    As a result, certain points that might otherwise appear bright become dark, and vice versa. This explains the formation of distinct interference patterns in many optical experiments.

    Understanding phase shifts at boundaries is therefore essential for analyzing interference effects in thin films and reflected light systems.

    Option d – π

    The central fringe in Newton’s rings observed in reflected light appears

    a) Black

    b) White

    c) Bright

    d) Colored

    Explanation:
    Newton’s rings are circular interference fringes formed when a plano-convex lens is placed on a flat glass plate. Between the curved surface of the lens and the plate, a thin air film of varying thickness is created. When monochromatic light falls on this system, part of the light is reflected from the upper surface of the air film and another part from the lower surface. These two reflected rays interfere with each other.

    At the exact point where the lens touches the glass plate, the thickness of the air film becomes essentially zero. The two reflected rays originate from surfaces with different optical properties. Because one reflection occurs from a boundary leading to a denser medium, it introduces a phase change of half a wavelength. The other reflected ray does not experience this same phase reversal.

    As a result, even though the path difference due to thickness is nearly zero at the center, the phase change causes destructive interference between the two reflected waves. Destructive interference means the intensity of light at that point becomes minimal.

    Therefore, the central point of the ring pattern appears dark when the pattern is observed in reflected light.

    Option a – Black

    The colorful appearance of oil films floating on water is mainly due to ( CSIR Net Physics Previous Year Question Papers MCQ )

    a) Light analysis

    b) Interference

    c) Diffraction

    d) Polarization

    Explanation:
    When a thin layer of oil spreads over the surface of water, it forms a film whose thickness is extremely small and varies slightly from place to place. When white light falls on this thin film, part of the light reflects from the top surface of the oil while another portion travels through the oil layer and reflects from the boundary between oil and water.

    These two reflected waves travel slightly different optical paths before reaching the observer. Because the thickness of the oil layer is comparable to the wavelength of visible light, the path difference between the reflected waves becomes significant. This difference allows the waves to interfere with each other.

    White light contains many wavelengths corresponding to different colors. For certain film thicknesses, some wavelengths interfere constructively and become more intense, while others undergo destructive interference and become weak or disappear. Since the thickness of the oil film changes across the surface, different wavelengths are reinforced at different locations.

    This selective reinforcement of particular wavelengths causes the bright, shifting colors commonly seen on oil films floating on water.

    Option b – Interference

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