WBCHSE Class 12 Physics Light Interference Questions and Answers

Light Wave And Interference Of Light Questions and Answers

Question 1. A point object t is placed on the axis of a convex lens at a distance greater than the focal length of the lens. What is the shape of the refracted wavefront?
Answer:

The wavefront of light rays, emitted born a point source it spherical convex. After being refracted through the Iens, the wavefront becomes spherical concave. This implies that after refraction through the lens, light rays converge

Question 2. What will be the nature of the wavefront of the direct sunlight and why will it be so?
Answer:

Plane wavefronts. Actually, the spherical wavefronts with the sun at the centre be have effective as plane wavefronts on the earth, at a very large distance from the sun.

Question 3. Interference fringe does not contradict the law of conservation energy – justify
Answer:

In an interference pattern, there is no loss or destruction of light energy in the dark fingers area. The energy just gets shifted from the region of the dark band to the region of the bright band. Total energy remains the same.It can be shown that the average intensity of a set of simultaneous consecutive dark and bright fringes is the same as the intensity of usual illumination in the same region.

Hence, interference fringe does not contradict the law of conservation of energy

WBBSE Class 12 Light Interference Q&A

Question 4. Two media of refractive indices μ₁ and μ₂ (μ₁ and μ₂ ) are separated by a plane surface. If some part of a plane wavefront is in the first medium and other port of the wavefront Is the second medium, show the shape of the wavefront In this position diagrammatically. stance colours reach here in the same phase and bri
Answer:

AB and BC are two plane wavefronts inclined to each other

Question 5. Why don’t two light sources of the same type produce an Interference pattern?
Answer:

The primary condition for the formation of interference fringes is that the two sources must be coherent and the phase difference between the intended waves for interference pattern on superposition, must remain constant

Extended light sources are aggregations of point sources where temperature varies from point to point which affects the radiation. Due to the random vibration of particles emitting light, there is a continuous change in phase. Thus no two-point sources from extended sources, maintain a constant phase difference and, hence are not coherent. Interference can be produced by waves from coherent sources only. Therefore a general illumination is found all over the screen rather than any interference fringes

Key Concepts in Light Interference Questions

Question 6. What are non-localised fringes?
Answer:

Interference fringes can be found on a screen placed anywhere in front of the coherent sources. Thus, the interference pattern is not localised in a region and hence these are called non-localised fringes

Question 7. Explain the change that occurs In the interference pattern in Young’s double slit experiment, if white light is used instead of monochromatic light
Answer: 

If white light is used instead of monochromatic light In Youngs’ double slit experiment, there will be bright-coloured fringes on either side of the white central fringe. White light of different coloured lights has different wavelengths Each of these waves produces a different characteristic interference pattern.

So the interference pattern looks coloured. But the line at the centre of this coloured. But the line at the centre of this interference coloured pattern remains at equal disfrom each of the coherent sources; hence light of all colours is here in the same phase and bright white light is produced.

Question 8. If all experiments related to Young’s double-slit Interference are performed underwater, what changes in
Answer:

The wavelength of light is smaller in water than in air. Since fringe width is proportional to the wavelength of light, fringe width will decrease and lines of interference pattern will be thinner

Question 9. Two waves whose Intensities are in the ratio 9: 1 interfere. Find the ratio of the intensities of bright and dark fringes.
Answer:

Let A₁ and A₂ be the amplitudes of two waves of intensities fj and J2 respectively

Given that \(\frac{I_1}{I_2}=\frac{9}{1}\)

Again \(\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}=\frac{9}{1}\)

Or, \(\frac{A_1}{A_2}=\frac{3}{1}\)

If A_max  and A_min are the amplitudes of bright and dark bands, then

⇒ \(\frac{A_{\max }}{A_{\min }}=\frac{A_1+A_2}{A_1-A_2} \text { or, } \frac{A_{\max }}{A_{\min }}=\frac{3+1}{3-1}\)

= \(\frac{4}{2}=\frac{2}{1}\)

⇒ \(\frac{I_{\max }}{I_{\min }}=\frac{A_{\max }^2}{A_{\min }^2}=\frac{4}{1}\)

Question 10. In Young’s double-slit experiment, one of the covered

1. With translucent e silts is paper and an opaque plate.
2. What changes will be observed In the Interference pattern in each case

Answer: 

1. An interference pattern will be formed and the fringe width will be unchanged. the difference in intensities of dark and bright fringes trill decreases i.e., bright fringes trill become less bright and dark fringers will become less dark.

2. Interference pattern trill vanishes and a continuous illumination trill be seen on the screen

Short Answer Questions on Young’s Double Slit Experiment

Question 11. What Is the effect on the interference pattern In Young’s double silt experiment if,

1. The screen is moved away from the silts
2. Separation between the slits is increased

Answer:

The width of interference fringes in Young’s double slit experiment, y = \(\frac{D}{2 d} \lambda\)

Here, D = distance of the screen from the slits,

2d = separation between the two slits,

λ = wavelength of light.

If the screen is moved away, D would increase, and so this fringe width would also increase.

If the separation 2d is increased, this fringe width will decrease

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Question 12. In Young’s double-slit experiment, if the distance between the two slits is halved and the distance between the screen a and plane of slits is doubled, how will the interference pattern be affected?
Answer: 

Fringe width, y = \(\frac{\lambda D}{2 d}\), 2d = separation between the slits, D = distance between screen and slits, A = wavelength of incident light wave.

Now if the distance between slits is d and the distance between the screen and the slit is changed to 2D then the fringe width will become,

⇒ \(\frac{\lambda D}{2 d}\)

Hence, the fringe width will become 4 times the earlier width.

Question 13. In Young’s double slit experiment, what is the path difference between the two light waves forming the 5th bright band on the screen?
Answer:

Path difference for the n-th bright band,

δ = 2n

\(\frac{\lambda}{2}\) = nλ

Given n = 5

So, δ = 5

Question 14. What is the significance of the optical path?
Answer:

The optical path is the distance travelled by light rays in a vacuum in the same time that it takes to traverse a certain distance (x) in a medium of refractive index. Its value is given by x.

Further, the change in phase for two rays of the same frequency will remain the same if they cover equal optical path lengths.

Question 15. Two light beams of intensities I and 4I, respectively form interference fringes on a screen. For the two beams, the phase difference at point A is \(\) and point B is. Find the difference in result intensities as A And B
Answer:

Resultant intensity at A,

\(I_A=I+4 I+2 \sqrt{I \cdot 4 I} \cos \frac{\pi}{2}\) = 5I

Resultant intensity at B,

⇒ \(I_B=I+4 I+2 \sqrt{I \cdot 4 I} \cos \pi=I\)

Hence, difference in resultant intensity = 5I – I = 4I

Question 16. The ratio of the amplitudes of two waves emitted from a pair of coherent sources is 2: 1. If the two waves superpose, what will be the ratio of the maximum and minimum intensities? What would have been the intensity at different points on the screen if the sources were not coherent?
Answer:

Let the amplitudes of the waves be A and A2 respectively.

By hypothesis

⇒ \(\frac{A_1}{A_2}=\frac{2}{1} \quad \text { or, } \frac{A_1+A_2}{A_1-A_2}=\frac{2+1}{2-1}=\frac{3}{1}\)

Or, \(\frac{A_{\max }}{A_{\min }}=\frac{3}{1}\)

⇒ \(\frac{I_{\max }}{I_{\min }}=\frac{A_{\max }^2}{A_{\min }^2}=\frac{9}{1}\)

Ratio of intensities =9:1.

For incoherent sources, at any point on the screen, the intensity

would be the sum of the intensities of two waves.

If amplitudes are 2A and A respectively then, intensities

7 = 4A² and 2 = A²

Resultant intensity at any point on screen = 4A²+ A² – 5

Common Questions on Interference Patterns in Light

Question 17. If the ratio of maximum to minimum intensities of the fringes, produced in Young’s double slit experiment is 4:1, what is the ratio of the amplitudes of light In a wave of coherent sources?
Answer: In this case

⇒ \(\frac{I_{\max }}{I_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}\)

Or, \(\frac{4}{1}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2} \text { or, } \frac{A_1+A_2}{A_1-A_2}=\frac{2}{1}\)

Or, \(\frac{A_1+A_2+A_1-A_2}{A_1+A_2-A_1+A_2}=\frac{2+1}{2-1}\)

Or, \(\frac{A_1}{A_2}=\frac{3}{1}\)

Question 18. Light waves of different intensities from two coherent sources superpose to interfere. If the ratio of the maximum intensity to minimum intensity is 25, find the ratio of the intensities of the sources.
Answer:

Here

⇒ \(\frac{I_{\max }}{I_{\min }}=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2}\)

Or, 25 = \(=\frac{\left(A_1+A_2\right)^2}{\left(A_1-A_2\right)^2} \quad \text { or, } \frac{A_1+A_2}{A_1-A_2}\) = 5

Or, \(\frac{A_1+A_2+A_1-A_2}{A_1+A_2-A_1+A_2}=\frac{5+1}{5-1}\)

Or, \(\frac{A_1}{A_2}=\frac{6}{4}=\frac{3}{2}\)

Or, \(\frac{I_1}{I_2}=\frac{A_1^2}{A_2^2}=\frac{9}{4}\)

I₁= I₂= 9:4

Question 19.  In a laboratory, interference fringes are observed In air medium. Tlie laboratory is now evacuated by removing air. If other conditions remain unaltered, what changes will be observed in the fringe pattern?
Answer:

The refractive index of air is slightly more than 1 . labora¬ tory being evacuated, the refractive index of the medium decreases. Hence wavelength increases. As fringe width is directly propor¬ tional to the wavelength of light used, fringes of marginally increased width will be observed

Question  20. In an Interference pattern by two identical slits, the intensity of the central maximum is I. What will be the intensity at the same spot if one of the slits is close

Let the amplitudes of the waves be a and a

a_max = a+a = 2a

So, I_max  = a²_max = 4a²

= 4I₀

[I₀ = Intensity due to each slit]

When one of the slits is closed, the intensity at the same spot is \(I_0=\frac{I_{\max }}{4}=\frac{I}{4}\)

Question 21. In a double-slit experiment, Instead of taking slits of equal width, one slit is made twice as wide as the other. Then how will the maximum and minimum intensities change?
Answer:

In case 0f interference of two waves with the same amplitude (a)

a_max = a+a = 2a;

Amin = a-a = 0

Maximum intensity ∝ a² : minimum intensity = 0

In case of interference of two waves with amplitudes a and A (A>a)

a_max = a+A: a_min = A – a ≠ 0

Minimum intensity  0 and maximum intensity (a+A)> 4 about

Hence, the maximum and minimum intenpattern will increase

Question 22. Monochromatic light of wavelength 589 nm is incident on a water surface from air What are the wavelength, frequency and speed of

1. Reflected and
2. Refracted light?

1. For reflected waves, the length and speed remain the same.

∴  \(\frac{c}{\lambda}=\frac{3 \times 10^8}{589 \times 10^{-9}}\) and c= 3 × 10⁸m .s^-1

∴  Frequency f = \(\frac{c}{\lambda}=\frac{3 \times 10^8}{589 \times 10^{-9}}\) = \(\)

= 5.09 × 10¹⁴ HZ

2. In the case of refraction only the frequency if fixed

= \(\frac{3 \times 10^8}{1.33}\)

= \(=\frac{v}{f}=\frac{2.20}{5.09 \times 10^{14}}\)

= 44 nm

Practice Questions on Coherent Sources in Interference

Question 23. What is the shape of the wavefront in each of the following cases?

1. Light diverges from a point source.
2. Light emerges out of a convex lens when a point source is placed at its focus.
3. The portion of the wavefront of light from a distant star intercepted by the earth. Pt

Answer:

1. Spherical
2. Plane
3. Plane

Question 24. In Young’s double slit experiment using monochromatic light of wavelength A, the intensity of light at a point on the screen where the path difference is A is k units. What is the intensity of light at a point where the path difference is
Answer: 

Resultant intensity at a point in Youngs double slit

⇒ \(I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi\)

When path difference = λ, phase difference

β = \(\frac{2 \pi}{\lambda} \cdot \lambda=2 \pi\)

∴ I +I’+2\(\sqrt{I \cdot I}\) = 4I

∴ k = 4I

When path difference = \(\frac{\lambda}{3}\) , phase difference

Φ= \(\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{3}=\frac{2 \pi}{3}\)

I” = I+I+2\(\sqrt{I \cdot I}\). cos \(\sqrt{I \cdot I} \cdot \cos \frac{2 \pi}{3}\) = I

I” = I = \(\frac{k}{4}\)

Question 25. In a double slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experiment apparatus is immersed in water? hike refractive Index of water to be \(\frac{4}{3}\)

Angular width, θ = \(\frac{\lambda}{d}\) = 0.2°

When the entire experiment is done underwater,\(=\frac{\lambda}{u}\)and angular width,

= \(\frac{\lambda^{\prime}}{d}=\frac{\lambda}{\mu d}\)

= \(\frac{\theta}{\mu}=\frac{0.2}{4 / 3}\)

= 0.15

Question 26. Use Huygens’ principle to show that a point object placed in front of a plane mirror produces a virtual image at the back of the mirror whose distance is equal to the distance of the object from the mirror.
Answer:

Let A be a point source of light at a distance y from the plane mirror MM’. In the absence of the mirror let the wavefront travel A’ in time t. But due to the presence of the M mirror the reflected wavefront will reach We point A in the same time interval t.

Thus AA’ = ct (here, c= velocity of light

i.e., AO+ OA’ = ct

Again, AO+OA = ct

OA = OA’

Important Definitions in Light Interference

Question 27. Following is a list of some factors which could possibly influence the speed of wave propagation. Nature of the source, The direction of propagation, Motion of the source and or observer, Wavelength, Intensity of the wave.

On which of these factors, if any, does:

1. The speed of light in a vacuum,
2. The speed of light in any medium like glass or water, depends.

Answer:

1. The speed of light in a vacuum is an absolute (universal) constant and does not depend on any factor.
2. The speed of light in any other medium depends only on the wavelength of light that the medium

Question 28. In a double slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1 0. What is the spacing between the two slits?
Answer: 

Using, dsinθ = nλ we get

d = \(\frac{n \lambda}{\sin \theta}=\frac{1 \times 600 \times 10^{-9}}{\sin 0.1^{\circ}}\)

= \(3.43 \times 10^{-4}\)m

Question 29.

1. Using Huygens’ principle, draw the diagrams to show the nature of the wavefronts when an incident plane wavefront gets

1. Reflected from a concave mirror,
2. Refracted from a convex lens.

2. Draw a diagram showing the propagation of a plane wavefront from a denser to a rarer medium and verify Snell’s law of refraction
Answer:

Let be the time taken by the wavefront to travel the distance BC, Thus BC = vt

Similarly AE. = v₂t

For triangles ABC And AEC,

sin i = \(\frac{B C}{A C}=\frac{v_1 t}{A C}\)

And sin r \(\frac{A E}{A C}=\frac{v_2 t}{A C}\)

Where i and r are the angles of incidence and refraction respectively

\(\frac{\sin i}{\sin r}=\frac{v_1}{v_2}\)…………… (1)

If c is the speed of light vacuum, then

⇒ \(\mu_1=\frac{c}{v_1}\)

Or, \(v_1=\frac{c}{\mu_1}\)

And \(\mu_2=\frac{c}{v_2}\)

Or, \(v_2=\frac{c}{\mu_2}\)

Here, μ₁₂ and μ₂ are known “as the refractive indices of medium 1 and medium 2 respectively. From equation

⇒ \(\frac{\sin i}{\sin r}=\frac{c}{\mu_1} \times \frac{\mu_2}{c}=\frac{\mu_2}{\mu_1}\)

Or, in i = sin

This is Snell’s law of refraction.

Examples of Applications of Light Interference

Question 30. A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interfer¬ ence fringes in Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are sepa¬ rated by 0.28 mm, calculate the least distance from /. the central bright maximum where the bright fringes of the two wavelengths coincide
Answer:

⇒ \(\frac{n_1 \lambda_1 D}{2 d}=\frac{n_2 \lambda_2 D}{2 d}\) [2d= distance between the slits

Or, \(\frac{n_1}{n_2}=\frac{\lambda_2}{\lambda_1}\)

= \(\frac{600}{800}=\frac{3}{4}\)

∴ 3rd order of 800 nm will overlap with 4th order of 600 nm

∴  Distance of the point from central bright maximum

y = \(\frac{n_1 \lambda_1 D}{2 d}=\frac{3 \times 800 \times 10^{-9} \times 1.4}{2.8 \times 10^{-4}}\)

= 12 × 10^-3 m

= 12 mm

Question 31. If one of two identical slits producing interference in Young’s experiment is covered with glass, so diet die light intensity passing through it is reduced to 50%, find the ratio ofdie maximum and minimum intensity of the fringe in the interference pattern.

What kind of fringes do you expect to observe if white light is used instead of monochromatic light?
Answer:

Let the amplitudes of the light waves passing through the slits be and a2 and the corresponding intensi¬ ties be  I₁  and I₂.

According to the die problem,

⇒ \(I_2=0.5 I_1=\frac{I_1}{2}\)

∴ \(a_2^2=\frac{a_1^2}{2}\)

Or, \(a_2=\frac{a_1}{\sqrt{2}}\)

= \(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_1+a_2\right)^2}{\left(a_1-a_2\right)^2}=\left(\frac{a_1+\frac{a_1}{\sqrt{2}}}{a_1-\frac{a_1}{\sqrt{2}}}\right)^2=\left(\frac{1+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}}}\right)^2\)

= \(\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)^2 \approx 34\).