AP Physics B - Multiple Choice Practice Questions on Physical Optics

“One of the deep secrets of life is that all that is really worth the doing is what we do for others.”

– Lewis Carroll

Today we shall discuss a few typical multiple choice questions on physical optics:

Questions  (1) and (2) are based on the following statement:

A beam of monochromatic light having wave length λ, frequency f and intensity I in air enters a glass slab of refractive index 1.5. After traveling through the slab, the beam of light emerges through the opposite face of the slab and passes through air.

(1) Inside the slab the wave length and frequency of the light are respectively 

(a) 1.5 λ and 1.5 f

(b) λ/1.5 and f

(c) 1.5 λ and f

(d) λ and f

(e)  λ and 1.5 f

The frequency of the light is unchanged where as the speed of the light is decreased within the slab. Note that the refractive index of a medium is the ratio of the speed of light in free space (or air at ordinary pressures) to the speed in the medium. Therefore, the speed of light (v) in the glass slab is given by

            v = c/1.5 where ‘c’ is the speed of light in free space (or air).

The wave length λ, frequency f and speed v are related as

            v = f λ  

Or,       λ = v/f

Since the speed in glass is (1/1.5) times the speed in air, the wave length in glass is (1/1.5) times the wave length in air [Option (b)].

(2) The energy of the photons in glass is

(a) 1.5 times the energy in air

(b) 1/1.5 times the energy in air

(c) the same as the energy in air

(d) 3 times the energy in air

(e) 1/3 times the energy in air

The energy E of a photon is given by

             E = hf where h is Planck’s constant and f is the frequency. Since the frequency of the light is the same in both air and glass, the energy of the photons in glass is the same as the energy in air [Option (c)].

(3) White light is passing through a transparent plastic slab. Inside the slab

(a) the green component  travels with maximum speed

(b) the green component  travels with minimum speed

(c) the violet component  travels with maximum speed

(d) the red component  travels with maximum speed

(e) all components travel with the same speed

The refractive index of any transparent medium is maximum for light of violet colour and minimum for light of red colour.

[This is why violet rays are deviated most and red rays are deviated least while traveling through a glass prism, producing dispersion of white light].

The speed of light (v) in the plastic slab is given by

            v = c/n where ‘c’ is the speed of light in free space (or air) and n is the refractive inex of the slab.

Since the refractive index is the least for light of red colour, it follows that red component  travels with maximum speed [Option (d)].

(4) In an experiment with Young’s double slit, a student measures the intensity of the central maximum of the interference pattern as I. If one of the slits is covered, what will be the intensity at the position of the central maximum?

(a) I/2

(b) I/4

(c) I/(√2)

(d) I

(e) 2I

When both slits are open, suppose the resultant amplitude (of light wave) at the position of the central maximum is a. If one of the slits is covered, the amplitude at the position of the central maximum becomes a/2.

Intensity is directly proportional to the square of the amplitude. Therefore we have

            I α a² and

            I₁ α (a/2)²

 Therefore I₁ = I/4, as given in option (b).

(5) Four laser sources produce light waves y₁, y₂, y₃, and y₄ given (with usual notations) by

             y₁ = a sin ωt

             y₂ = a sin 2ωt

             y₃ = 2a sin (ωt + φ)

             y₄ = 2a sin (3ωt + φ)                                                               

Superposition of which two waves can produce interference fringes?

(a) y₁ and y₂

(b) y₂ and y₃

(c) y₃ and y₄

(d) y₁ and y₄

(e) y₁ and y₃

Waves y₁ and y₃ have the same wave length and hence they can produce interference fringes.

[The angular frequencies of y₁ and y₃ are equal].

The correct option is (e).

You will find a few useful multiple choice questions in this section here.

AP Physics B- Multiple Choice Questions on Interference and Diffraction

In the post dated 6^th January 2008, the essentials of wave optics required for you to work out multiple choice questions were discussed. (You will find that post by clicking on the label ‘wave optics’ below this post). In that post, the following question was given for your practice:

A student prepared a Young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. [For smoking, the plate was held above the flame of a kerosene lamp]. Accidentally he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the intensity at the interference maximum to that at the interference minimum?

(a) 4

(b) 2

(c) 3

(d) 6

(e) 9

As promised, I give the solution here:

The ratio of intensities of light proceeding from the slits, I₁: I₂ = 4 : 1

Therefore, the ratio of amplitudes of light proceeding from the slits,

a₁:a₂ = √I₁: √I₂ = 2:1

The amplitudes at the maximum and minimum of the interference pattern are therefore in the ratio (a₁+a₂) : (a₁–a₂) = 3:1

The intensities at the maximum and the minimum are in the ratio3²:1² = 9:1 [Option (e)].

[The intensity is proportional to the square of the amplitude].

We will discuss a few more typical multiple choice questions in this section:

(1) A single slit diffraction pattern is obtained on a screen using yellow light. If the yellow light is replaced by blue light without making any other changes in the experimental set up, what will happen to the diffraction bands?

(a) Bands will disappear

(b) Bands will become broader and farther apart

(c) Bands will become broader and crowded together

(d) Bands will become narrower and farther apart

(e) Bands will become narrower and crowded together

The angular width of the central maximum is 2λ/a and the minima are located at θ = ± λ/a, ± 2λ/a, ± 3λ/a…etc.

This shows that when the wave length is decreased as is the case when blue light is used instead of yellow light, the bands will become narrower and crowded together [Option (e)].

(2) If white light is used in Young’s double slit experiment, what will happen to the interference bands?

(a) No bands will be obtained

(b) Many bands will be obtained as in the case of monochromatic light, but they will be coloured except the centre of the central band which will be white

(c) Very few bands will be obtained, but they will be coloured except the centre of the central band which will be white

(d) Many bands will be obtained as in the case of monochromatic light, but all of them will be white

(e) Very few bands will be obtained, but all of them will be white.

The centre of the central bright band (n = 0) will be white since all colours will satisfy the condition of zero path difference (and therefore of brightness). Since the condition for the first (order n =1) bright band is satisfied, at the point nearest to the centre, by the shortest wave length (violet), the first bright band will be coloured with the edge nearest to the centre violet and the edge farthest to the centre red. The remaining bright bands also will be coloured likewise. There will be very few distinct bands since the same point on the screen will satisfy the condition for brightness for one wave length and the condition for darkness for another.

The correct option therefore is (c).

(3) In an arrangement for Young’s double slit experiment, the separation between the slits is 1 mm. It is found that 8 bands of the double slit interference pattern can occupy the central maximum of the single slit diffraction pattern produced by one of the slits. What is the wjdth of each slit?

(a) 0.2 mm

(b) 0.25 mm

(c) 0.3 mm

(d) 0.35 mm

(e) 0.4 mm

The angular fringe width in the case of the double slit interference pattern is λ/d so that the total angular separation of 8 interference bands is 8λ/d. Since the angular width of the central maximum in the case of the single slit diffraction pattern is 2λ/a, we have

8λ/d = 2λ/a.

Putting d = 1 mm, a = d/4 = 0.25 mm.

[You should note that in Young’s double slit experiment, the pattern you see on the screen is actually a superposition of single-slit diffraction from each slit, and the double-slit interference pattern. Therefore, there will be a broader diffraction peak in which several fringes of smaller width due to double-slit interference appear. The angular width of a double slit intetrference fringe is λ/d where as the angular width of the central maximum due to the single slit diffraction is 2λ/a. Therefore, the number of interference fringes occurring in the broad central diffraction maximum depends on the ratio d/a, that is the ratio of the distance between the two slits to the width of a slit. If ‘a’ is very small, the diffraction pattern will become very flat and we will observe the two-slit interference pattern with an appreciable number of interference maxima of almost equal intensity in the central region].

You will find some useful multiple choice questions in this section at physicsplus: Multiple Choice Questions on Wave Optics.

AP Physics B- Optics: Equations to be Remembered in Interference and Diffraction

If I have seen a little further it is by standing on the shoulders of Giants.

–Sir Isaac Newton

In Physical Optics the topics included are: (i) Interference and diffraction (ii) Dispersion of light and the electromagnetic spectrum. These topics carry 5% of the total points.

You must remember the following equations to make you strong in answering multiple choice questions involving interference and diffraction:

(1) Refractive index(n) of medium 2 with respect to medium 1 is given given by

n = n₂/ n₁ = λ₁/ λ₂ = v₁/v₂ where λ₁ and λ₂ are the wave lengths and v₁ and v₂ are the velocities of light in medium 1 and medium 2 respectively].

(2) The resultant displacement produced (at a particular point in a medium) by a number of waves is the vector sum of the displacements produced by each of the waves. For instance, if two waves arriving at a point P produce displacements y₁ and y₂, the resultant displacement at P is the vector sum of y₁ and y₂. If the interfering waves have amplitudes a₁ and a₂ and they arrive at the point P with a phase difference of φ, the resultant amplitude at P will be [a₁²+ a₂²+ 2a₁ a₂ cosφ]^1/2, in accordance with vector addition. If the interfering waves have intensities I₁ and I₂ the resultant intensity (I) at P will be given by

I = I₁ + I₂ + [2√(I₁I₂)] cosφ, since the intensity is directly proportional to the square of the amplitude.

If φ = 0, 2π, 4π, 6π,…..(an even multiple of π), the intensity is maximum (I_max) given by

I_max= I₁ + I₂ + [2√(I₁I₂)]

If φ = π, 3π, 5π, 7π,…..(an odd multiple of π),, the intensity is minimum (I_min) given by

I_min= I₁ + I₂ – [2√(I₁I₂)]

(3) Two sources of light (or any wave) are said to be coherent if they produce waves of the same frequency with a constant phase difference between them. (The condition of same frequency is implied in the condition of constant phase difference; yet it is usually specified when we define coherent sources). If two coherent sources produce waves of equal amplitude (a) and therefore equal intensity (I₀), then the resultant intensity (I) when they interfere is given by

I = I₀ + I₀ + [2√(I₀I₀)]cosφ = 2I₀(1+cosφ)

If the phase difference between the interfering waves (of equal intensity) is zero or an even multiple of π (in other words, if the path difference between the waves is zero or an integral multiple of the wave length λ), the intensity is maximum (I_max) given by

I_max = 2I₀(1+1) = 4I₀ …..(Path difference = nλ where n = 0,1,2,3,4,…..etc)

If the phase difference between the interfering waves (of equal intensity) is an odd multiple of π (in otherwords, if the path difference between the waves is an odd multiple of half wave length), the intensity (I_min) is zero given by

I_min = 2I₀(1–1) = 0……(Path difference = (2n+1)λ/2 where n = 0,1,2,3,4,….etc)

(4) In the interference fringe pattern produced by Young’s double slit, the distance of the n^th bright fringe (x_n) from the central (zero order) bright fringe is given by

x_n = nλD/d, where n = 0, ±1, ±2, ±3….etc.

where λ is the wave length of light, D is the distance between the screen (on which the fringes are formed) and the double slit and ‘d’ is the separation between the two slits S₁ and S₂ forming the double slit (fig.).

The distance of the n^th dark fringe (x_n) from the central (zero order) bright fringe is given by

x_n = (2n+ 1) λD/2d, where n = 0, ±1, ±2, ±3….etc. [We use ± sign for the integers to indicate the fringes on both sides of the central fringe]

The fringe width β is the distance between the centres of consecutive brigt fringes or dark fringes and is given by

β = x_n+1 –x_n

Therefore, β = λD/d

Angular fringe width is independent of the distance of the screen and is given by

β_θ = β/D = λ/d

(5) When light traveling through a rarer medium gets reflected from the surface of a denser medium, it suffers a phase change of π, which is equivalent to a path change (path difrference) of λ/2

(6) Colours in thin transparent films can be observed in reflected light as well as transmitted light. The condition for brightness in the reflected light is

2μtcosr = (2n+ 1)λ/2 where μ is the refractive index of the film, t is its thickness, r is the angle of refraction in the film and n = 0,1,2,3,4,….etc.

[A film of negligible thickness cannot satisfy the above condition and hence it cannot appear bright in reflected light. (It will appear dark)].

The condition for brightness in transmitted light is

2μtcosr = nλ where n = 0,1,2,3,4,….etc.

[A film of negligible thickness can satisfy the above condition (for n = 0) and hence it will appear bright in transmitted light].

You should note that Newton’s rings in reflected light are produced by the interference between the light waves reflected from the two surfaces of an air film contained in between a glass plate and a lens.

(7) In the diffraction pattern produced by a single slit, the angular separation between the first minima is 2λ/a where a is the width of the slit. The angular width of the central maximum is therefore 2λ/a.

The centre of the central maximum is at O (fig.). On either side of O, positions of minimum intensity are obtained when the angle θ = ±nλ/a where n = 1,2,3…etc.

Therefore, the minima are located at θ = ± λ/a, ± 2λ/a, ± 3λ/a…etc. (Fig.) Positions of

secondary maxima are obtained on either side of O corresponding to

θ = ± (2n+ 1)λ/2a where n = 1,2,3….etc.

Therefore, the secondary maxima are located at θ = ±3λ/2a, ±5λ/2a, ±7λ/2a….etc. The intensity of the secondary maxima goes on decreasing. [The first secondary maximum has intensity less than 5% of the intensity of the central maximum].

(8) In a plane diffraction grating, the n^th principal maximum is obtained at an angle θ given by

(a+b) sinθ = nλ where a is the width of the slit and b is the width of the opaque region between the slits. [a+b is called the grating element]

In the next post questions on wave optics will be discussed. Here is a question which you must be able to work out in no time if you have understood the phenomenon of interference.

A student prepared a Young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. [For smoking, the plate was held above the flame of a kerosene lamp]. Accidentally he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the intensity at the interference maximum to that at the interference minimum?

(a) 4

(b) 2

(c) 3

(d) 6

(e) 9

Work out the above question. I’ll be back shortly with the solution and of course, more solved questions.