Solution of S Chand PhysicsClass 10. We covered all the Solution of S Chand PhysicsClass 10 in this post for free so that you can practice well for the exam.
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In a biprism experiment, 60 interference fringes are obtained using sodium Light of wavelength 6000 Å. With the experimental setup unchanged, sodium Light is replaced by mercury Light filtered to a wavelength of 5000 Å. How many fringes will now be observed?
a) 60
b) 65
c) 72
d) 77
Explanation: The question focuses on how the number of interference fringes changes in a biprism setup when the wavelength of the Light source is altered while keeping all other experimental conditions constant. In interference phenomena, fringe formation depends on the wavelength of Light, coherence, and geometrical arrangement. When the setup remains unchanged, the number of observable fringes is inversely related to the wavelength used.
In this case, two different monochromatic Light sources are considered with different wavelengths. Since fringe width in a biprism experiment is directly proportional to wavelength, a decrease in wavelength leads to a reduction in fringe width. When fringe width becomes smaller, more fringes can fit within the same observation region. This means the total number of visible fringes increases when a shorter wavelength source is used.
To reason through the problem, compare the two wavelengths provided. The first case uses a longer wavelength, while the second uses a shorter one. The ratio between these wavelengths determines how the fringe pattern scales. The experimental geometry does not change, so only the wavelength ratio affects the outcome. The final observation is obtained by applying this proportional relationship conceptually to adjust the original number of fringes accordingly.
A useful analogy is compressing a SET of equally spaced lines into a smaller spacing area, allowing more lines to appear within the same width.
Option c – 72
In a biprism arrangement, the separation between successive bright fringes is 0.32 mm when red Light of wavelength 6400 Å is used. What will be the change in this separation if blue Light of wavelength 4800 Å is employed under identical conditions?
a) 0.24 mm
b) 0.16 mm
c) 0.80 mm
d) 0.08 mm
Explanation: This question deals with how fringe separation changes in a biprism interference experiment when the wavelength of Light is modified while keeping all other parameters constant. Fringe separation, also called fringe width, depends on wavelength, distance between virtual sources, and screen distance. When geometry remains unchanged, wavelength becomes the only varying factor affecting the pattern.
In interference, fringe width is directly proportional to wavelength. A higher wavelength produces wider spacing between fringes, while a lower wavelength results in narrower spacing. Since the experiment switches from a longer wavelength red Light to a shorter wavelength blue light, the fringe spacing will reduce. The change is governed purely by the ratio of wavelengths, as the optical setup remains fixed.
To analyze the situation, compare how the second wavelength relates to the first. Because the new wavelength is smaller, the fringe pattern compresses proportionally. This means each bright and dark fringe moves closer to its neighbors. The change in separation is determined by scaling the original fringe width according to this proportional reduction.
A helpful way to visualize this is imagining evenly spaced markers on a stretched string. When the string is shortened without changing marker count, the spacing between markers decreases uniformly.
Option d – 0.08 mm
During a biprism experiment, the fringe width increases by 25% when the distance between the slit and the focal plane of the eyepiece is increased by 20 cm. What was the original distance between the slit and the eyepiece?
a) 60 cm
b) 80 cm
c) 100 cm
d) 120 cm
Explanation: The question examines how fringe width in a biprism interference setup varies when the distance between the slit and the observation system is changed. Fringe width depends on the wavelength of light, the distance between the virtual coherent sources, and the distance between the slits and the observation screen or eyepiece. When all other factors remain constant, the fringe width varies directly with the slit-to-screen distance.
Here, the system experiences a proportional increase in fringe width due to an increase in the effective observation distance. Since the relationship is linear, any percentage change in fringe width corresponds to the same percentage change in this distance parameter. The added distance value is given, which allows reconstructing the original setup dimension using proportional reasoning.
To solve conceptually, the increase in distance causes a proportional increase in fringe width. This relationship helps establish an equation between original and final distances using percentage change. By interpreting the added separation as part of the final configuration, the original slit-to-eyepiece distance can be inferred through subtraction after scaling.
A simple analogy is a projector casting stripes on a wall: moving the wall farther apart spreads the stripes more widely in direct proportion.
Option b – 80 cm
In a biprism experiment, the distance between the third and the sixth dark fringes on the same side of the central bright fringe is measured as 4.5 mm. What is the fringe width?
a) 2 mm
b) 1 mm
c) 1.5 mm
d) 3 mm
Explanation: This question is based on determining fringe width using positional differences between dark fringes in an interference pattern. In a biprism experiment, dark and bright fringes are equally spaced when conditions are ideal. The fringe width represents the distance between two successive dark or bright fringes and remains constant across the pattern.
Here, the distance between two non-adjacent dark fringes is given. The third and sixth dark fringes are separated by a fixed number of fringe intervals. Each interval corresponds to one fringe width. Therefore, the total measured distance includes multiple equal spacings. By identifying how many steps lie between the two mentioned fringes, the fringe width can be determined.
The reasoning involves counting the number of gaps between the specified fringes and relating total distance to repeated equal separations. Since the system is uniform, dividing the total distance by the number of intervals gives the fringe width. This method is commonly used in experiments to reduce measurement error by using widely separated fringes instead of adjacent ones.
An analogy is measuring steps between floor tiles: instead of one tile gap, counting multiple tiles improves accuracy in determining single spacing.
Option c – 1.5 mm
The fringe width in a biprism setup is 0.32 mm when red light of wavelength 6400 Å is used. If blue light of wavelength 4000 Å is used with the same arrangement, what is the percentage change in fringe width?
a) 62.5%
b) 50%
c) 37.5%
d) 27.5%
Explanation: This question explores how fringe width changes when the wavelength of light in a biprism interference experiment is altered while keeping the setup unchanged. Fringe width depends directly on wavelength, so any change in wavelength produces a proportional change in fringe separation.
In this situation, the system switches from a longer wavelength red light to a shorter wavelength blue light. Since fringe width is directly proportional to wavelength, reducing wavelength reduces fringe width proportionally. The percentage change is determined by comparing the ratio of the new wavelength to the original wavelength and interpreting the difference relative to the initial value.
The reasoning involves recognizing that all geometric parameters remain constant, so only wavelength scaling affects the result. A smaller wavelength compresses the fringe pattern, reducing spacing between consecutive bright or dark fringes. The percentage change is then derived from how much smaller the new wavelength is compared to the original one.
A helpful analogy is shrinking a printed barcode horizontally: reducing the scale compresses all bars uniformly, decreasing spacing between them in the same proportion.
Option c – 37.5%
If the distance between the slit and the eyepiece in a biprism experiment is increased by 20%, while all other parameters remain unchanged, how will the fringe width change?
a) Decrease by 20%
b) Increase by 20%
c) Increase by 10%
d) Remain unchanged
Explanation: The question examines how the fringe width in a biprism interference setup responds when only the distance between the slit and the observation system is increased, while all other experimental factors such as wavelength and slit separation remain constant. Fringe width in interference depends directly on the distance between the virtual source plane and the observation screen or eyepiece.
Since fringe width is directly proportional to this distance, any change in it produces a corresponding proportional change in fringe spacing. When the distance increases, the interference pattern spreads out because light rays have more space to diverge before reaching the observation plane. This leads to a uniform scaling of the entire fringe pattern.
To reason through the problem, the percentage increase in distance directly translates into the same percentage change in fringe width because all other variables are fixed. The pattern does not distort; it only expands or contracts uniformly. Thus, the change in fringe width can be determined purely through proportional reasoning.
A simple analogy is stretching a printed pattern on a rubber sheet: pulling the sheet outward increases spacing between all printed lines uniformly.
Option b – Increase by 20%
For obtaining clear and distinct interference fringes in a biprism experiment, the refracting edge of the biprism should be:
a) Approximately parallel to the broad slit
b) Exactly parallel to the broad slit
c) Perpendicular to the narrow slit
d) Exactly parallel to the narrow slit
Explanation: This question focuses on the proper alignment required in a Fresnel biprism experiment to obtain a stable and visible interference pattern. In such experiments, coherent light sources are produced by splitting a single slit into two virtual coherent sources using a biprism. The quality of interference fringes depends heavily on correct orientation of the optical components.
For clear fringes, the refracting edge of the biprism must be aligned in a way that ensures symmetrical division of the incoming light from the slit. Proper alignment ensures that the two virtual sources formed are coherent, of equal intensity, and properly spaced for observable interference. If the orientation is incorrect, the overlap of light waves becomes irregular, reducing fringe visibility or producing distorted patterns.
The reasoning is based on achieving maximum symmetry in wavefront division so that constructive and destructive interference occurs uniformly across the screen. Any angular misalignment reduces coherence quality and fringe contrast. Therefore, correct orientation is essential for sharp and evenly spaced fringes.
A helpful analogy is splitting a beam of light through a perfectly centered prism: only precise alignment ensures both split beams remain balanced and overlap properly on the screen.
Option d – Exactly parallel to the narrow slit
In a biprism experiment, for which colour of light is the fringe width maximum?
a) Violet
b) Green
c) Yellow
d) Red
Explanation: The question relates to how fringe width depends on the wavelength of light used in a biprism interference experiment. Fringe width is directly proportional to wavelength when all other experimental conditions remain constant, such as slit separation and distance to the observation screen.
Different colors of visible light have different wavelengths. Among them, red light has the longest wavelength while violet has the shortest. Since fringe width increases with wavelength, the color with the largest wavelength produces the widest spacing between interference fringes.
The reasoning involves comparing wavelengths across the visible Spectrum. Longer wavelength means greater separation between successive bright and dark fringes because the phase difference changes more slowly over distance. Therefore, the fringe pattern becomes more spread out for longer wavelengths.
A simple analogy is drawing waves on a rope: longer waves create wider spacing between peaks, while shorter waves produce tightly packed oscillations.
Option d – Red
What is the effect on the interference fringes when the refracting angle of the biprism is increased from 0° to 179°?
a) Fringes move closer together
b) Fringes spread farther apart
c) No change in the fringe pattern
d) Fringes cease to appear
Explanation: This question explores how changing the refracting angle of a biprism affects the interference pattern formed in a Fresnel biprism experiment. The biprism works by creating two virtual coherent sources through refraction, and the angle of the prism controls the deviation of light rays and thus the separation between these virtual sources.
As the refracting angle increases significantly, the deviation of light rays also increases. This leads to a larger separation between the virtual coherent sources. Since fringe width is inversely proportional to the separation between sources, increasing this separation reduces fringe width and brings fringes closer together.
In extreme cases, if the angle becomes too large, the overlap region of the two wavefronts may reduce or disappear. Without sufficient overlap, interference cannot occur effectively, and the visibility of fringes decreases or vanishes entirely.
The reasoning is based on wave overlap conditions: interference requires overlapping coherent waves. Increasing angular deviation disrupts this overlap geometry.
A helpful analogy is two flashlights slowly moving apart: as they diverge, the overlapping bright region shrinks until no overlapping pattern is visible.
Option b – Fringes spread farther apart
While determining the separation between two virtual sources in a biprism experiment, the eyepiece is moved away from the slit so that its distance from the slit is:
a) Less than four times the focal length of the lens
b) Less than twice the focal length of the lens
c) Equal to the focal length of the lens
d) Greater than four times the focal length of the lens
Explanation: This question relates to the experimental condition required in a Fresnel biprism setup when measuring the separation between the two virtual coherent sources formed by the biprism. The eyepiece or observation system must be positioned in a way that allows clear resolution of the interference fringes and accurate measurement of fringe geometry.
In such optical setups, the clarity and measurability of fringes improve when the observation point is placed sufficiently far from the source slit. This ensures that the overlapping wavefronts have enough distance to interfere properly and form well-defined, steady fringes. If the eyepiece is too close, the pattern may not be fully developed or may be difficult to analyze due to insufficient wave overlap.
The reasoning is based on the requirement for proper spatial development of interference patterns. A larger slit-to-eyepiece distance allows the wavefronts to spread and overlap more uniformly, making the fringe system more distinct and easier to measure accurately. This condition is essential for precise determination of virtual source separation using geometric relationships.
A simple analogy is watching ripples on water: standing farther away gives a clearer view of overlapping wave patterns compared to observing them too close to the source.
Option d – Greater than four times the focal length of the lens
In Fresnel’s biprism experiment, if the separation between the two narrow slits is doubled, what adjustment must be made to the distance D between the slits and the screen to keep the fringe spacing unchanged?
a) D
b) 2D
c) D/2
d) 4D
Explanation: This question focuses on maintaining constant fringe width in a Fresnel biprism interference setup when one of the key parameters, namely the separation between the two virtual or effective slits, is changed. Fringe width depends on wavelength, distance to the screen, and slit separation.
Fringe width is directly proportional to the distance between the slits and the screen, and inversely proportional to the separation between the slits. If slit separation increases, fringe width decreases unless compensated by adjusting another parameter. To maintain the same fringe spacing, the system must preserve the overall proportional balance between these quantities.
When the slit separation is doubled, the fringe width would naturally reduce to half if no other change is made. To counteract this effect, the distance between the slits and the screen must be adjusted proportionally so that the ratio remains constant. This ensures that the interference pattern spacing does not change.
The reasoning is based on maintaining a constant ratio between geometric parameters in the interference equation. Adjusting distance is the only way to compensate for increased slit separation while preserving fringe width.
A simple analogy is balancing a seesaw: if weight increases on one side, adjusting distance on the other side restores equilibrium.
Option b – 2D
In Fresnel’s biprism experiment, what happens if one of the slits is covered with an opaque black sheet?
a) Fringes become broader
b) Fringes become less bright
c) Fringes become narrower
d) Interference fringes disappear
Explanation: This question deals with the fundamental requirement for interference in a Fresnel biprism experiment, which is the presence of two coherent light sources. The biprism creates two virtual sources from a single slit, and these sources must emit waves that overlap to produce an interference pattern.
If one of the effective slits is blocked using an opaque material, one of the two coherent wave contributions is removed. Without two overlapping wavefronts, there can be no constructive and destructive interference. As a result, the characteristic bright and dark fringe pattern disappears.
The reasoning is based on the principle that interference requires at least two coherent waves of similar frequency and phase relationship. Removing one source eliminates the necessary condition for superposition, leaving only uniform illumination without variation in intensity.
In practical terms, the screen would no longer show alternating bright and dark bands, but instead a uniform light distribution from the remaining source.
A simple analogy is clapping with one hand: without two interacting elements, the intended effect (sound pattern or rhythm) cannot be produced.
Option d – Interference fringes disappear
What happens to the interference pattern if the biprism is rotated about a horizontal axis?
a) Fringe width decreases
b) Fringe width increases
c) Bright and dark fringes exchange positions
d) Interference fringes vanish
Explanation: This question explores how the orientation of a biprism affects the interference pattern in a Fresnel biprism experiment. The biprism works by splitting light from a single slit into two virtual coherent sources, and proper alignment is essential for stable interference fringes.
When the biprism is rotated about a horizontal axis, the geometry of the refracting surfaces relative to the incident light changes. This alters the direction in which the light rays are deviated, which in turn affects the overlap region of the two emerging wavefronts. If the overlap is reduced or disturbed, the interference pattern becomes unstable.
As the rotation increases, the condition for symmetric wavefront division is lost. This leads to a reduction in fringe visibility and may eventually cause the interference pattern to disappear if coherent overlap is no longer maintained. Even slight misalignment can distort fringe spacing or reduce contrast between bright and dark bands.
The reasoning is based on the requirement that both virtual sources must remain coherent and properly aligned for stable interference. Any angular change that disrupts symmetry affects fringe formation.
A simple analogy is projecting two overlapping slides: tilting one slightly can blur or misalign the combined image, reducing clarity.
Option d – Interference fringes vanish
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