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The position, relative size and nature of the image formed by a concave lens for an object placed at infinity are respectively
(a) at focus, diminished and virtual
(b) at focus, diminished and real
(c) between focus and optical centre, diminished and virtual
(d) between focus and optical centre, magnified and real
Explanation: When an object is at infinity, Light rays approaching a lens are parallel. Concave lenses diverge parallel rays, making them appear to come from a focal point on the same side as the object. Images formed by concave lenses are always virtual, upright, and smaller than the object.
For an object at infinity, the parallel rays after passing through the concave lens appear to diverge from the focal point, so the image forms between the optical center and the focal point. The rays do not actually converge, which makes the image virtual. Since the rays are spreading, the image appears diminished compared to the object.
Imagine sunlight passing through a concave lens: it seems to originate from a tiny virtual point near the lens rather than focusing on a screen.
Thus, for distant objects, concave lenses always form small, virtual, and upright images near the focal point.
Option a – at focus, diminished and virtual
The upper and lower portions in a common type of bi-focal lenses are respectively
(a) concave and convex
(b) convex and concave
(c) both concave of different focal lengths
(d) both convex of different focal lengths
Explanation: Bifocal lenses are designed to help with both distance and near vision in a single lens. They consist of two distinct segments: one for distance and one for near objects. Concave lenses correct nearsightedness, and convex lenses correct farsightedness.
In typical bifocals, the upper portion is for distance vision and is concave to handle myopia, while the lower portion is convex to magnify close objects for hypermetropia or presbyopia. The curvature of each section corresponds to the focal length required for its specific vision correction.
Think of a window with two panels: the top portion shows distant scenery clearly, while the lower portion enlarges a book held close.
This arrangement allows smooth switching between near and far vision without changing glasses.
Option a – concave and convex
In the phenomenon of dispersion of Light, the Lightwave of the shortest wavelength is
(a) accelerated and refracted the most
(b) slowed down and refracted the most
(c) accelerated and refracted the least
(d) slowed down and refracted the least
Explanation: Dispersion occurs when white Light splits into its component colors while passing through a prism. Each color has a different wavelength, and the amount of refraction depends on the wavelength: shorter wavelengths bend more, longer wavelengths bend less.
Since the shortest wavelength Light (violet) experiences the greatest change in speed when entering a denser medium, it is refracted the most. Longer wavelengths, like red, bend less. The refractive index varies slightly with wavelength, which is why each color emerges at a slightly different angle, producing the visible Spectrum.
This explains why rainbows appear with violet at one end and red at the other; the colors separate according to their wavelengths.
In summary, dispersion separates colors because shorter wavelengths bend more strongly than longer wavelengths.
Option a – accelerated and refracted the most
Wavelengths of which of the following colour of the visible Spectrum of Light are maximally absorbed by green plants?
(a) Green and Yellow
(b) Red and Blue
(c) Green and Red
(d) Blue and Yellow
Explanation: Plants absorb Light for photosynthesis primarily using pigments like chlorophyll. Chlorophyll absorbs red and blue wavelengths most efficiently, while green light is largely reflected, giving leaves their green color.
Red light has longer wavelengths (~620–700 nm) and blue light has shorter wavelengths (~450–495 nm), both of which carry energy suitable for driving photosynthesis reactions. Green and yellow light are less absorbed because chlorophyll reflects them.
This selective absorption is why plants appear green to our eyes. Sunlight passing through a canopy has its red and blue components absorbed, maximizing photosynthetic activity.
Thus, red and blue wavelengths are the most strongly absorbed by green plants for energy conversion.
Option b – Red and Blue
Consider the following statements 1. A person with myopia can see distant objects distinctly but cannot see nearby objects clearly. 2. A person with hypermetropia cannot see distant objects clearly. 3. A person with presbyopia can see nearby objects without corrective glasses. Which of the statements given above is/are not correct?
(a) 1, 2 and 3
(b) 1 and 2
(c) 1 and 3
(d) Only 3
Explanation: Myopia, hypermetropia, and presbyopia affect vision differently. Myopia is nearsightedness, making distant objects blurry. Hypermetropia is farsightedness, affecting near vision but usually distant objects are seen clearly. Presbyopia occurs with age as the lens loses flexibility, reducing near vision clarity.
Understanding each condition is crucial. Myopia causes difficulty seeing far; hypermetropia makes close tasks hard; presbyopia affects near vision even if distance vision remains good. Analyzing the statements requires comparing these characteristics with what each description says.
In essence, identifying incorrect statements involves knowing which conditions affect near versus distant vision and how the lens’ shape or flexibility contributes.
This can be compared to camera lenses: a fixed-focus lens struggles with close objects, while a diverging lens struggles with distant ones.
Thus, evaluating the statements requires careful matching of eye conditions to their visual effects.
Option a – 1, 2 and 3
Which one of the following statements is not correct?
(a) The longest wavelength of the light visible human eye is about 700 mm
(b) The shortest wavelength of light visible to the human eye is about 400 mm
(c) The wavelength of gamma rays is longer than that of X-rays
(d) The ability of a telescope to form a separable image of close objects is called its resolving power
Explanation: Light has a range of visible wavelengths (~400–700 nm). Gamma rays and X-rays have much shorter wavelengths than visible light, far beyond human perception. A telescope’s resolving power is its ability to distinguish two close objects.
The statements involve comparing wavelength ranges and understanding telescope function. Gamma rays never have longer wavelengths than X-rays, and visible light has limits for human vision. Knowledge of the electromagnetic Spectrum helps identify incorrect claims.
Think of the electromagnetic Spectrum as a ruler of wavelengths: visible light occupies a small section between UV and infrared, while X-rays and gamma rays are far shorter.
The incorrect statement can be identified by comparing the wavelength ranges of different types of radiation.
Option c – The wavelength of gamma rays is longer than that of X-rays
If the image of an object, formed by a concave mirror is virtual, erect and magnified, then the object is placed
(a) the principal focus
(b) at the centre of curvature
(c) beyond the centre of curvature
(d) between the pole of the mirror and the principal focus
Explanation: Concave mirrors can form real or virtual images depending on the object’s position relative to the focal point. When the object is between the mirror’s pole and the focal point, the reflected rays diverge, and the image appears virtual and upright behind the mirror.
Magnification occurs because the reflected rays appear to come from a larger image. If the object is beyond the focal point, the image becomes real and inverted. Understanding ray diagrams for concave mirrors is key to visualizing this.
It’s like holding a hand near a spoon: close to the reflective surface, your face appears upright and larger, whereas farther away it appears inverted.
Thus, the virtual and magnified image occurs only when the object is near the mirror, between the pole and focal point.
Option d – between the pole of the mirror and the principal focus
The dispersion process forms a Spectrum due to white light on the prism. The light wave with the shortest wavelength falls a
(a) refracts the most
(b) does not change the path
(c) refracts the least
(d) is reflected by the side of the prism
Explanation: Dispersion separates white light into colors because each wavelength refracts differently in a prism. Shorter wavelengths slow down more in a denser medium, resulting in greater bending. Longer wavelengths bend less.
Violet light, with the shortest wavelength (~380–450 nm), deviates the most, while red light bends the least. The prism’s refractive index varies with wavelength, which causes the angular spread of colors and forms the visible Spectrum.
This is similar to sunlight passing through raindrops to form a rainbow: shorter wavelengths appear at one end, longer at the other.
Overall, shorter wavelength light is refracted more due to its interaction with the prism material.
Option a – refracts the most
A ray of white light strikes the surface of an object. If all the colours are reflected the surface would appear
(a) black
(b) white
(c) grey
(d) opaque
Explanation: White light contains all visible wavelengths. The color perceived depends on which wavelengths are absorbed or reflected by a surface. If all colors are reflected equally, the combination appears white.
If certain wavelengths are absorbed while others are reflected, the object takes on the reflected color. Surfaces that absorb all colors appear black. Reflection of all wavelengths produces maximum light intensity to the eyes.
This is why a white paper reflects most of the sunlight, making it appear bright and white compared to a black surface.
In short, reflecting all wavelengths equally results in a white appearance to the observer.
Option b – white
A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is
(a) -30 cm
(b) – 20 cm
(c) – 40 cm
(d) – 60 cm
Explanation: Magnification relates the size of the image to the size of the object. For mirrors, magnification is given by:
M = v / u = image height / object height
where v is the image distance and u is the object distance.
Since the image is smaller than the object, magnification is less than one, indicating a real and inverted image. The mirror formula connects object distance, image distance, and focal length:
1 / f = 1 / u + 1 / v
The signs of u and v depend on the mirror convention (real images are positive in front of the mirror).
As an analogy, imagine holding a small object near a spoon: as you move it closer or farther, the size of its reflection changes. This change in image size is mathematically linked to the focal length through magnification.
Overall, the focal length of a concave mirror is determined by the distances of the object and image and their relative sizes, using the mirror formula.
Option b – – 20 cm
The angle of incidence i and refraction r are equal in a transparent slab when the value of i is
(a) 0º
(b) 45°
(c) 90°
(d) depends on the material of the slab
Explanation: In a transparent slab, refraction follows Snell’s law: n₁ sin i = n₂ sin r, where n₁ and n₂ are the refractive indices of the media. The angle of incidence equals the angle of refraction when light strikes perpendicular to the surface.
At normal incidence, the light passes straight through without bending because sin i = sin r = 0. This is independent of the slab’s material, as refraction depends on the sine of the angles.
It’s similar to looking straight into water: if you look straight down, your view is not bent; only angled lines appear displaced.
Thus, equal angles of incidence and refraction occur when light enters perpendicular to the slab surface.
Option a – 0º
The refractive index of a transparent medium is greater than one because
(a) Speed of light in vacuum
(b) speed of light in a transparent medium
(c) Speed flight in vacuum=speed of light in a transparent medium
(d) Frequency of light wave changes when it moves from rarer to denser medium
Explanation: Refractive index n is defined as n = c / v, where c is the speed of light in vacuum and v is the speed in the medium. Since light slows down in a denser medium (v < c), the refractive index is always greater than 1. The frequency of light remains constant across media, but the speed and wavelength decrease in a denser medium, causing refraction. For analogy, it’s like running from a smooth surface to sand: your speed decreases, so the effective “index” of the sand is higher than the smooth surface. In summary, the refractive index is >1 because light travels slower in the medium than in vacuum.
Option b – speed of light in a transparent medium
You are given three media A, B and C of refractive index 1.33, 1.65 and 1.46. The medium in which the light will travel fastest is
(a) A
(b) B
(c) C
(d) equal in all three media
Explanation: The speed of light in a medium is v = c / n, where n is the refractive index. Higher refractive index means slower speed.
Comparing n values: A = 1.33, B = 1.65, C = 1.46. The smallest n corresponds to the highest speed. Therefore, light travels fastest in the medium with the lowest refractive index.
Think of it like cars moving through different terrains: smoother terrain (lower n) allows higher speed.
Thus, among the three media, light travels fastest in the one with the lowest refractive index.
Option b – B
Light from the Sun falling on a convex lens will converge at a point called
(a) centre of curvature
(b) focus
(c) the radius of curvature
(d) optical centre
Explanation: A convex lens is a converging lens. Parallel rays from distant sources like the Sun bend towards the principal axis and meet at a point called the focal point.
The distance from the lens to this point is the focal length. This principle is used in magnifying lenses, cameras, and Solar concentrators.
An analogy is sunlight focused by a magnifying glass to form a small bright spot, which is the focal point.
Hence, parallel light rays from the Sun converge at the lens’s focus.
Option b – focus
Which of the following statements is/are true?
(a) A convex lens has 4 dioptre power having a focal length of 0.25 m
(b) A convex lens has -4 dioptre power having a focal length of 0.25 m
(c) A concave lens has 4 dioptre power having a focal length of 0.25 m
(d) A concave lens has – 4 dioptres having a focal 0.25 m
Explanation: Lens power (P) is related to focal length (f) in meters as P = 100 / f (cm) or P = 1 / f (m). Positive power corresponds to convex lenses, negative to concave lenses.
Understanding sign conventions and formulas is essential to determine whether a lens is concave or convex, its focal length, and power. Using the relationship between power and focal length helps in verifying correctness of any statement about a lens.
For analogy, lens power is like the magnifying strength of a glass; positive magnifies, negative diverges.
Thus, assessing lens statements requires comparing the given focal length and sign with the formula for power.
Option a – A convex lens has 4 dioptre power having a focal length of 0.25 m
Magnification produced by a rearview mirror fitted in vehicles
(a) is less than one
(b) is more than one
(c) is equal to one
(d) can be more than or less than one depending upon the position of the object in front of it.
Explanation: Rearview mirrors are usually convex. Convex mirrors always produce virtual, erect, and diminished images regardless of object distance. The magnification is always less than one because the reflected rays diverge, making the image appear smaller.
This allows drivers to see a wider field of view behind the vehicle, even though the image is smaller.
Think of looking into a spoon from the outside curve: your reflection appears smaller but shows a wider area.
In short, convex rearview mirrors reduce magnification but expand the visible area.
Option a – is less than one
As the sunlight passes through the Atmosphere, the rays are scattered by tiny particles of dust, pollen, soot and other minute particulate Matter present there. However, when we look up, the sky appears blue during mid-day because
(a) blue light is scattered most
(b) blue light is absorbed most
(c) blue light is reflected most
(d) ultraviolet and yellow components of sunlight combine
Explanation: Scattering of light depends on wavelength. Rayleigh scattering causes shorter wavelengths (blue) to scatter more than longer wavelengths (red).
During mid-day, sunlight passes through a smaller thickness of the Atmosphere, so blue light is scattered in all directions, giving the sky its blue color. Red and yellow light scatter less, which is why the Sun itself looks yellowish.
This is similar to the effect of a prism: shorter wavelengths bend more and are more noticeable.
Hence, the sky appears blue because blue light is scattered most strongly.
Option a – blue light is scattered most
A lens has a power of +2.0 Dioptre. Which one of the following statements about the lens is true?
(a) The lens is concave and has a focal length of 0.5 metres
(b) The lens is convex and has a focal length of 2.0 metre
(c) The lens is convex and has a focal length of 0.5 metres
(d) The lens is concave and has a focal length of 2.0 metre
Explanation: Lens power (P) is related to focal length (f) in meters: P = 1 / f. Positive power indicates a convex lens, and negative indicates concave.
Given +2 D, focal length f = 1 / 2 = 0.5 m. Convex lenses converge light to a point and have positive power, so a +2 D lens is convex with 0.5 m focal length.
It’s like a magnifying glass: a converging lens brings rays together, focusing at a point determined by its focal length.
Thus, positive power lenses are convex, and their focal length is the reciprocal of the power in meters.
Option c – The lens is convex and has a focal length of 0.5 metres
Which of the following lenses will bend the light rays through the largest angle?
(a) A lens with power +2.0 D
(b) Lens with power +2.5 D
(c) Lens with power-1.5 D
(d) Lens with power-2.0 D
Explanation: Lens power determines the degree to which light is converged or diverged. Higher absolute power bends light more. Positive power bends light to converge; negative power diverges light.
Among lenses, the one with the largest absolute power causes the maximum bending of light rays, independent of sign. Convex or concave, the bending depends on the magnitude of the power.
Think of it as sharper or weaker magnifying glasses: the stronger lens bends rays more sharply.
Hence, the lens with the largest absolute power produces the greatest deviation of light.
Option b – Lens with power +2.5 D
The nature of the image formed by a concave mirror when the object is placed between the focus (F) and centre of curvature (C) of the mirror observed by us is
(a) real, inverted and diminished
(b) virtual, erect and smaller in size
(c) real, inverted and enlarged
(d) virtual, upright and enlarged
Explanation: Concave mirrors form different images depending on object location. When an object is between F and C, reflected rays converge beyond C, producing a real, inverted, and enlarged image.
Ray diagrams show that rays from the object intersect after reflection, creating a magnified image on the same side as the object for a real image. Inversion occurs because the top and bottom rays cross after reflection.
It is similar to focusing sunlight through a concave mirror: an object slightly beyond the focus appears magnified on a screen.
Thus, the image is real, inverted, and larger when the object is between F and C.
Option c – real, inverted and enlarged
The nature of the image formed by a convex mirror when the object’s distance from the mirror is less than the distance between the pole and focal point (F) of the mirror would be
(a) real, inverted and diminished in size
(b) real, inverted and enlarged in size
(c) virtual, upright and diminished in size
(d) virtual, upright and enlarged in size
Explanation: Convex mirrors are diverging mirrors. No Matter where the object is placed, the reflected rays diverge and appear to come from a point behind the mirror.
This produces a virtual, erect, and diminished image. Even if the object is close to the mirror, the image remains virtual and upright because diverging rays never converge in front of the mirror.
It’s similar to the outer surface of a spoon: holding an object near it shows a smaller upright reflection.
Thus, convex mirrors always form virtual, upright, and diminished images regardless of object distance.
Option c – virtual, upright and diminished in size
If a man’s face is 25 cm in front of a concave shaving mirror producing an erect image 1.5 times the size of his face, the focal length of the mirror would be
(a) 75 cm
(b) 25 cm
(c) 15 cm
(d) 60 cm
Explanation: Magnification (M) relates image height to object height: M = v / u, where v is image distance and u is object distance. Positive magnification indicates an erect image.
Using the mirror formula: 1 / f = 1 / u + 1 / v. Substituting the known values of u and M gives the focal length. For virtual, erect images, the image appears on the same side as the object.
It’s like holding your face near a spoon: the erect magnified image appears larger and closer to the mirror than the actual distance.
Thus, focal length can be calculated using the object distance and magnification with the mirror formula.
Option a – 75 cm
As light travels from a rarer to a denser medium it will have
(a) increased velocity
(b) decreased velocity
(c) decreased wavelength
(d) both (b) and (c)
Explanation: When light passes from a rarer to a denser medium, its speed decreases (v decreases), causing the wavelength to reduce while frequency remains constant.
The decrease in velocity and wavelength results in bending towards the normal (refraction). This is consistent with Snell’s law: n₁ sin i = n₂ sin r.
Think of a car moving from tarmac onto sand: the speed drops, and the distance between “bumps” along its path shortens, like light’s wavelength decreasing.
In summary, both velocity and wavelength decrease when light enters a denser medium.
Option d – both (b) and (c)
Which of the following statements is NOT correct regarding the travel of sound waves?
Explanation: sound waves are mechanical and require a medium to propagate. They can travel through Solids, liquids, and gases, but cannot travel through vacuum because there are no particles to transmit vibrations.
This contrasts with electromagnetic waves, which can travel in vacuum. Knowledge of wave types is essential to identify exceptions or incorrect statements.
It’s similar to trying to send a ripple through empty space — nothing moves because no medium exists.
Option d – sound waves can travel through a vacuum
Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become
(a) double
(b) half
(c) four times
(d) one fourth
Explanation: Radiation pressure is proportional to the intensity of light and the area exposed: F = (I / c) × A.
If the area is reduced by half while intensity remains constant, the force exerted on the surface also reduces by half. Total reflection doubles the momentum transfer per photon, but the dependence on area remains linear.
Imagine sunlight falling on a sheet: halving the sheet area halves the push exerted by light.
Therefore, radiation force is directly proportional to the illuminated area.
(A) directly as the distance between the slit and the screen
(B) inversely as the wavelength
(C) directly as the width of the slit
(D) independent of the distance between the slit and the screen
Explanation: The fringe width (β) in single-slit Diffraction is given by β = λL / a, where λ is the wavelength, L is the distance between slit and screen, and a is slit width.
Fringe width increases with distance from slit to screen and decreases with larger slit widths or smaller wavelength. This explains why small slits produce wider fringes and longer light wavelengths produce slightly wider fringes.
It is analogous to water waves passing through a narrow opening: the narrower the opening, the more the waves spread.
Thus, Diffraction fringe width depends directly on distance and wavelength, inversely on slit width.
Option a – directly as the distance between the slit and the screen
Explanation: Scattering occurs when light interacts with small particles or molecules in a medium, causing the light to deviate in different directions. Rayleigh scattering explains why shorter wavelengths scatter more.
It differs from Diffraction (bending around edges), interference (overlapping waves), or dispersion (separation by wavelength). Scattering depends on particle size relative to wavelength.
An analogy is sunlight passing through dust-laden air: tiny particles scatter light, making the sky bright or causing colored sunsets.
Thus, scattering arises from interaction of light with small particles.
While both light and sound show wave character, Diffraction is much harder to observe in light. This is because
(A) light does not require a medium
(B) the wavelength of light is far smaller
(C) waves of light are transverse
(D) the speed of light is far greater.
Explanation: Diffraction depends on wavelength relative to obstacle size. Light has extremely small wavelengths (~10-7 m), so noticeable Diffraction occurs only with very small openings comparable to wavelength.
sound has much larger wavelengths (~10-1 m), making Diffraction around obstacles like doors easily observable.
It’s similar to water waves versus tiny ripples: small ripples barely bend around obstacles.
Hence, light’s tiny wavelength makes diffraction harder to observe in everyday scenarios.
Option b – the wavelength of light is far smaller
The compact disc shows color in white light due to the:
(A) interference
(B) polarization
(C) diffraction
(D) scattering
Explanation: A compact disc has closely spaced tracks forming a diffraction grating. When white light strikes it, different wavelengths interfere constructively or destructively at different angles, producing colorful patterns.
This is diffraction combined with interference: the spacing of tracks separates light by wavelength. Reflection alone does not produce the Spectrum.
Similar to a rainbow of colors appearing on a CD surface when tilted in sunlight.
Thus, the colorful pattern is a result of diffraction and interference of light waves.
Option c – diffraction
The bending of light around corners of an obstacle is called :
(A) dispersion
(B) refraction
(C) deviation
(D) diffraction
Explanation: When light encounters an obstacle or slit comparable in size to its wavelength, it bends around the edges. This phenomenon is called diffraction.
Refraction, reflection, dispersion, and deviation involve bending at interfaces or separating colors, not around corners. Diffraction explains why waves spread after passing through slits or around edges.
An analogy is water waves spreading after passing a small gap in a barrier.
Thus, bending of light around obstacles is explained by diffraction.
Option d – diffraction
A diffraction pattern is obtained using red light. Which of the following happens if the red light is replaced by blue light?
(A) There is no change
(B) Diffraction bands become narrower and more crowded
(C) Diffraction bands become broader and farther
(D) Bands disappear
Explanation: Diffraction fringe width is proportional to wavelength: β = λL / a. Blue light has a shorter wavelength than red, so replacing red with blue reduces fringe width.
Narrower fringes appear more closely spaced, and the overall diffraction pattern becomes denser. This effect is purely due to wavelength differences; the slit and screen distance remain constant.
It’s like squeezing waves in water: shorter waves spread less when passing through a gap.
Hence, using blue light makes diffraction bands narrower and more crowded than with red light.
Option b – Diffraction bands become narrower and more crowded
A parallel beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is :
(A) 2π
(B) π
(C) π/2
(D) π/3
Explanation: In single-slit diffraction, the first minimum occurs when path difference between rays from the slit edges equals one wavelength. Phase difference (φ) is related to path difference: φ = 2π × (path difference / λ).
At the first minimum, the path difference = λ, so φ = 2π. This phase difference causes complete destructive interference at the first minimum.
Think of two synchronized swimmers: if one completes a full cycle ahead of the other, their peaks and troughs cancel at certain positions.
Thus, the phase difference between the rays at the first minimum is 2π.
Option a – 2π
Diffraction of light was discovered by :
(A) Young
(B) Hertz
(C) Grimaldi
(D) Malus
Explanation: Diffraction is the bending of light around obstacles or through slits. Francesco Maria Grimaldi first systematically studied and described this effect in the 17th century.
He observed fringes formed when light passed through narrow openings and coined the term “diffraction” from Latin “diffringere,” meaning to break into pieces.
An analogy is water waves spreading after passing through a small gap: Grimaldi’s experiments highlighted similar behavior in light.
Thus, diffraction of light was first discovered and documented by Grimaldi.
Option c – Grimaldi
The fringe pattern observed in young’s double slit experiment is :
(A) diffraction pattern
(B) interference pattern
(C) a combination of diffraction and interference pattern
(D) neither interference nor diffraction pattern
Explanation: Young’s double slit experiment demonstrates interference of light. Two coherent sources produce overlapping waves, resulting in bright and dark fringes on a screen.
Bright fringes occur due to constructive interference (in-phase), and dark fringes occur due to destructive interference (out-of-phase). This pattern is called an interference pattern, distinct from single-slit diffraction patterns, although some diffraction effects are superimposed.
It’s similar to overlapping ripples in water: peaks and troughs combine to form alternating high and low regions.
Hence, the observed fringe pattern is an interference pattern.
Option c – a combination of diffraction and interference pattern
Yellow light is used in a single slit diffraction experiment with a slit width of 0.6 mm. If the yellow light is replaced by X-rays, then the observed pattern will reveal :
(A) that the central maximum is narrower
(B) more fringes will be seen
(C) less number of fringes will be seen
(D) no diffraction pattern will be seen
Explanation: Diffraction occurs when slit width is comparable to wavelength. Yellow light (~580 nm) is comparable to slit width, producing visible fringes.
X-rays have extremely small wavelengths (~10-10 m), much smaller than 0.6 mm. The diffraction effect becomes negligible, and no visible diffraction pattern appears.
It’s like trying to pass tiny ripples through a huge opening: the waves travel straight without noticeable spreading.
Thus, replacing yellow light with X-rays results in no observable diffraction pattern.
Option d – no diffraction pattern will be seen
The condition for observing Fraunhofer diffraction from a single slit is that the light wavefront incident on the slit should be :
(A) spherical
(B) plane
(C) cylindrical
(D) elliptical
Explanation: Fraunhofer diffraction requires plane (parallel) wavefronts so that rays reaching the slit are effectively parallel. This ensures the far-field diffraction pattern is clear and well-defined.
Spherical or curved wavefronts produce Fresnel diffraction instead, observed in near-field conditions. The plane wave approximation simplifies calculations for slit width, wavelength, and fringe position.
Think of it like parallel water waves hitting a barrier: far-field spreading is predictable.
Hence, plane wavefronts are required for Fraunhofer diffraction.
Option b – plane
The condition for observing Fresnel’s diffraction from a single slit is that the light wavefront incident on the slit should be :
(A) spherical
(B) plane
(C) cylindrical
(D) (A) or (C)
Explanation: Fresnel diffraction occurs when the source or screen is at a finite distance from the slit. The incident wavefront must be curved, either spherical or cylindrical.
This causes near-field diffraction, where rays interfere before reaching the far-field. Fresnel diffraction explains phenomena like shadow edge fringes or patterns near apertures.
An analogy is waves near a small obstacle in a pond: the curvature of wavefronts produces localized interference patterns.
Thus, Fresnel diffraction requires spherical or cylindrical wavefronts incident on the slit.
Option d – (A) or (C)
Sunlight filtering through tree leaves often makes circular patches on the ground because
(A) the sun is round
(B) light is transverse in nature
(C) the space through which light penetrates is round
(D) diffraction effects
Explanation: Each small gap between leaves acts as a pinhole, producing circular images of the Sun on the ground via rectilinear propagation of light.
This is similar to a pinhole camera: light travels straight from the source, forming inverted or circular images corresponding to the light source shape.
Hence, the circular patches correspond to the Sun’s shape as seen through small apertures.
Option d – diffraction effects
From Brewster’s law for polarization, it follows that the angle of polarization depends upon
(A) the wavelength of light
(B) the plane of vibration’s orientation
(C) the plane of polarization orientation
(D) all of these
Explanation: Brewster’s law states that reflected light is fully polarized when the reflected and refracted rays are perpendicular. The angle of polarization depends on refractive indices: tan θₚ = n₂ / n₁.
It is independent of wavelength for ideal surfaces, but real materials may show slight dependence. Orientation of the plane of vibration does not change this angle.
An analogy is shining light on a glass surface at a special angle to eliminate glare; this angle is determined by the refractive indices.
Thus, the angle of polarization is determined by the medium properties.
Option a – the wavelength of light
An unpolarised beam of transverse waves is one whose vibrations :
(A) are confined to a single plane
(B) occur in all directions
(C) occur in all directions perpendicular to their direction of motion
(D) has not passed through a polarised disk
Explanation: In unpolarized transverse waves, vibrations occur in all directions perpendicular to the wave propagation. Polarization restricts these vibrations to a single plane.
Electromagnetic waves like light are naturally unpolarized but can be polarized using filters or reflection.
Think of a rope shaken in multiple directions: all orientations are present until constrained.
Hence, unpolarized waves vibrate in all directions perpendicular to the motion.
Option c – occur in all directions perpendicular to their direction of motion
In the propagation of electromagnetic waves, the angle between the direction of propagation and the plane of polarization is :
(A) 0°
(B) 45°
(C) 90°
(D) 120°
Explanation: Electromagnetic waves consist of electric and magnetic fields oscillating perpendicular to each other and to the direction of propagation. The plane of polarization is defined by the orientation of the Electric Field.
Since the Electric Field is perpendicular to the wave propagation, the angle between the direction of propagation and the plane of polarization is 90°. This perpendicular relationship is a key property of transverse electromagnetic waves.
An analogy is a wave on a rope moving horizontally: the Oscillation of the rope is vertical, perpendicular to the motion direction.
Thus, the Electric Field orientation ensures a 90° angle with propagation.
Option a – 0°
Which of the following cannot be polarized?
(A) Radio waves
(B) Ultraviolet rays
(C) X-ray
(D) Ultrasonic waves
Explanation: Polarization occurs only in transverse waves, where oscillations are perpendicular to the direction of wave propagation. Longitudinal waves, like ultrasonic waves in air, oscillate along the propagation direction.
Hence, longitudinal waves cannot be polarized because restricting vibration to a plane is meaningless. Electromagnetic waves like X-rays or radio waves are transverse and can be polarized.
It’s like a slinky compressed and released along its length: you cannot confine its vibrations to a plane perpendicular to propagation.
Explanation: Longitudinal waves have oscillations along the direction of propagation. Polarization is a property of transverse waves because it involves restricting vibrations to a single plane perpendicular to propagation.
Reflection, refraction, and diffraction can occur for longitudinal waves, but polarization cannot. This is a fundamental difference between transverse and longitudinal waves.
An analogy is sound waves: no Matter how you try, you cannot filter their compressions into a single plane.
Thus, polarization is not observed in longitudinal waves.
Option c – polarization
Two polarizing plates have polarizing directions parallel so as to transmit maximum intensity of light. Through what angle must either plate be turned if the intensities of the transmitted beam are to drop by one-third that of the polarised beam?
(A) 54° 44′
(B) 125° 16′
(C) Both (A) and (B)
(D) 60°
Explanation: Malus’s law states that transmitted intensity through a polarizer is I = I₀ cos²θ, where θ is the angle between initial and analyzer polarization directions.
To reduce intensity to 2/3 of original: I/I₀ = cos²θ = 2/3. Solving gives θ ≈ 35°, which is the rotation required to achieve the intensity drop.
An analogy is tilting polarized sunglasses slightly: the transmitted light diminishes depending on the angle relative to light polarization.
Thus, the angle of rotation is determined by the cosine squared relationship.
Option c – Both (A) and (B)
A ray of light falls on a transparent glass plate, part of it is reflected and part is refracted. The reflected and refracted rays can be perpendicular to each other for :
(A) No angle of incidence
(B) Angle of incidence 90°
(C) More than one angle of incidence
(D) Only one angle of incidence
Explanation: Brewster’s law states that when reflected and refracted rays are perpendicular, reflected light is fully polarized. The angle of incidence θᵢ satisfies tan θᵢ = n₂ / n₁.
There is only one specific angle that satisfies this condition for a given pair of media. This ensures the reflected and refracted rays are at 90° to each other.
It’s like tilting a glass at the Brewster angle to eliminate glare: reflected and refracted rays are perpendicular.
Thus, the perpendicular condition occurs at only one specific angle of incidence.
Option d – Only one angle of incidence
Which of the following properties exhibit that light waves are transverse in nature? :
(A) interference
(B) diffraction
(C) polarization
(D) all of these
Explanation: Transverse waves oscillate perpendicular to propagation. Light shows interference, diffraction, and polarization—behaviors typical of transverse waves.
Polarization directly demonstrates transverse nature, as restricting oscillations to a plane is only possible for transverse waves. Interference and diffraction also depend on perpendicular oscillations interacting across wavefronts.
An analogy is a rope wave: only transverse waves can show interference or polarization patterns.
Hence, all listed properties indicate that light is transverse.
Option c – polarization
The fact that light is a transverse wave phenomenon derives its evidential support from the observation that :
Explanation: Polarization of light provides direct evidence for its transverse nature, since only transverse waves can be constrained to oscillate in a single plane.
Other phenomena like interference and diffraction can occur in both wave types, but polarization is exclusive to transverse waves.
For analogy, consider shaking a rope in one direction: the motion can be confined to a single plane, demonstrating its transverse nature.
Thus, light’s ability to be polarized proves it is transverse.
Option c – light shows polarizing effects
The phenomenon which does not take place in a sound wave is :
(A) polarization
(B) scattering
(C) interference
(D) diffraction
Explanation: Sound is a longitudinal mechanical wave. While reflection, interference, and diffraction occur, polarization does not, as it requires oscillations perpendicular to propagation.
Sound vibrations are along the wave’s travel direction, so restricting oscillations to a plane is impossible.
An analogy is trying to polarize compressions in a slinky: it cannot be done.
Hence, polarization does not occur in sound waves.
Option a – polarization
When a plane polarised light is passed through an analyzer and the analyzer is rotated through 90°, the intensity of emerging light :
(A) varies between maximum and zero
(B) varies between maximum and minimum
(C) becomes zero
(D) does not vary
Explanation: According to Malus’s law: I = I₀ cos²θ, rotating the analyzer changes the angle θ between the light’s polarization and analyzer axis.
At θ = 0°, intensity is maximum; at θ = 90°, intensity drops to zero. The intensity varies continuously between maximum and zero as the analyzer rotates.
An analogy is rotating polarized sunglasses: turning the glasses changes the brightness depending on alignment with light polarization.
Thus, intensity varies between maximum and zero.
Option a – varies between maximum and zero
The oscillating electric and magnetic Vectors of an electromagnetic wave are oriented along
(A) the same direction but differ in phase by 90°
(B) the same direction and are in phase
(C) mutually perpendicular directions and are in phase
(D) mutually perpendicular directions and differ in phase by 90°
Explanation: In an EM wave, the Electric FieldE and magnetic field B are perpendicular to each other and to the direction of propagation, forming a mutually perpendicular SET of Vectors.
They are in phase, meaning peaks of E coincide with peaks of B. This is a defining property of electromagnetic waves.
It’s like three mutually perpendicular axes (x, y, z) representing E, B, and propagation directions.
Thus, E and B Vectors are perpendicular to each other and to wave propagation, and in phase.
Option c – mutually perpendicular directions and are in phase
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