Two Questions (MCQ) from Wave Optics

Here are two questions from wave optics. The first question is meant for testing your understanding of coherent sources and interference while the second one is meant for testing your knowledge of the interference pattern produced by Young’s double slit.

(1) Two coherent sources emitting light of wave length λ and amplitude A produce an interference pattern on a screen. The maximum intensity of the interference pattern is I. If the light sources are not coherent but the wave length and amplitude are λ and A respectively, the maximum intensity on the screen will be

(a) I/4

(b) I/2

(c) I

(d) √I

(e) 2√I

At the interference maximum the amplitudes of the light waves get added. Therefore, the resultant amplitude at the interference maximum is A + A = 2A.

The intensity is directly proportional to the square of the amplitude so that we have

I α 4A²

Or, I = k×4A²……………(i)

where k is the constant of proportionality

Since the amplitude of each light wave is A, the intensity (I₀) of light produced by each source is given by

I₀ = k×A²

This gives I₀ = I/4 from (i).

Each source will produce uniform intensity I₀ everywhere on the screen and the resultant intensity everywhere will be 2I₀ (since there are two sources) and we have

2I₀ = 2×I/4 = I/2

The correct option is (b).

(2) The angular width of the interference fringes obtained in a double slit experiment using light of wave length λ is found to be θ. If the entire experimental arrangement is immersed in water having refractive index 4/3, the angular fringe width will be

(a) θ/4

(b) θ/3

(c) θ

(d) 4θ/3

(e) 3θ/4

The angular fringe width is λ/d where d is the separation between the slits. Therefore we have

θ = λ/d.

When the arrangement is submerged in water of refractive index n, the wave length of light is reduced to λ' given by

λ'= λ/n = 3λ/4 since n = 4/3

The angular fringe width now becomes θ'= λ'/d = 3λ/4d = 3θ/4.

You will find some useful multiple choice questions in this section at AP Physics Resources: Multiple Choice Questions on Interference and Diffraction .

AIEEE 2009 Multiple Choice Questions on Optics

I believe in standardising automobiles, not human beings

– Albert Einstein

The following three questions were included from optics in the All India Engineering/Architecture Entrance Examination (AIEEE) 2009 question paper:

(1) A transparent solid cylindrical rod has a refractive index of 2/√3. It is surrounded by air. A light ray is incident at the mid point of one end of the rod as shown in the figure. The incident angle θ for which the light ray grazes along the wall of the rod is

(1) sin^–1(1/√3)

(2) sin^–1(1/2)

(3) sin^–1(√3/2)

(4) sin^–1(2/√3)

If n is the refractive index and θ_c is the critical angle of the material of the rod, we have n = 1/sinθ_c so that

sinθ_c = 1/n = √3/2.

If r represents the angle of refracion (as shown in the figure), we have

1×sinθ = nsin r = (2/√3) ×sin(90º – θ_c)

Or, sinθ = (2/√3) ×cos θ_c

But cos θ_c = ½ since sin θ_c = √3/2

[You may imagine a right angled triangle with opposite side √3 and hypotenuse 2. The adjacent side will be 1]

Substituting for cos θ_c, we obtain

sinθ = (2/√3) × ½ = 1/√3 so that θ = sin^–1(1/√3)

(2) In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using the same scale for the two axes. A straight line passing through the origin and making an angle of 45° with the x-axis meets the experimental curve at P. The coordinates of P will be

(1) (4f, 4f)

(2) (2f, 2f)

(3) (f/2, f/2)

(4) (f, f)

This is a well known experiment which most of you might have performed. As such most of you know the answer: (2f, 2f)

We have 1/v – 1/u = 1/f

Applying the Cartesian sign convention, v is positive and u is negative so that

1/v + 1/u = 1/f

Since the straight line makes an angle of 45º, we have v = u at P and each must be equal to 2f to satisfy the equation, 1/v + 1/u = 1/f

(3) A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is

(1) 776.8 nm

(2) 393.4 nm

(3) 885.0 nm

(4) 442.5 nm

The fringe width β is given by

β = λD/d where λ is the wave length of light, D is the distance between the double slit and the screen and d is the double slit separation.

Since three fringes due to the light of wave length 590 nm exactly occupy the space occupied by four fringes due to the unknown light, we can write

3×590×D/d = 4λD/d where λ is the wave length of the unknown light (in nanometer).

This gives λ = 442.5 nm

You will find additional multiple choice questions (with solution) at the following locations:

1. apphysicsresources(i)

2. aphysicsresources(ii)

IIT-JEE 2008 Question (MCQ) on Young’s Double Slit

Today we will discuss two questions on Young’s double slit. The following multiple correct answers type question appeared in IIT-JEE 2008 question paper:

In a Young’s double slit experiment, the separation between the two slits is d and the wave length of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).

(A) If d = λ, the screen will contain only one maximum.

(B) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen.

(C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase.

(D) If the intensity of light falling on slit 2 is reduced so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase

The path difference [S₂N = (S₂P – S₁P) in fig.] between the interfering beams is d sinθ. Therefore for obtaining a maximum at the point P on the screen the condition to be satisfied is

dsinθ = nλ

When d = λ, sinθ = n

Since sinθ ≤ 1, n ≤ 1

The possible values of n are 0 and 1. But the maximum corresponding to n = 1 cannot be observed since θ cannot be 90º. Therefore, the central maximum of order zero corresponding to zero path difference (n = 0) alone can be observed. Option A is therefore correct

If λ < d < 2λ, the limiting values of d are λ and 2λ. In addition to the usual central maximum, fringes of order up to n satisfying the equation, dsinθ = nλ will be obtained, where d < 2λ.

If d = 2λ, we have 2λ = nλ/sinθ so that n = 2 sinθ.

Since sinθ should be less than 1 for observing the fringe, the maximum value of n must be 1. Therefore, the central maximum and the maximum of order n = 1 will be observed. Option B too is therefore correct

Options C and D are obviously incorrect since the intensity of the dark fringes will become zero when the light beams passing through the slits are of equal intensity.

Now consider another question on Young’s double slit:

Suppose the wave length λ and the double slit separation d in a Young’s double slit experiment are such that the 6^th dark fringe is obtained at point P shown in the above figure. The path difference (S₂P – S₁P) will be

(a) 5 λ

(b) 5 λ/2

(c) 6 λ

(d) 3 λ

(e) 11 λ/2

The dark fringe of order 1 (1^st dark fringe) is formed when the path difference is λ/2. The dark fringe of order 2 is formed when the path difference is 3λ/2 (and not 2λ/2) and the dark fringe of order 3 is formed when the path difference is 5λ/2 (and not 3λ/2). Generally, the dark fringe of order n is formed when the path difference is (2n – 1)λ/2. The 6^th dark fringe is therefore formed when the path difference is 11λ/2.

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You will find useful posts on wave optics here at apphysicsresources.