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 Practice Questions on Geometric Optics

"I know not with what weapons World War III will be fought, but World War IV will be fought with sticks and stones."

–Albert Einstein

Questions on optics were discussed on earlier occasions on this site. You can access those questions by clicking on the label ‘optics’ below this post. Generally simple questions will be asked from this section. But to answer them correctly you need to have a thorough knowledge of basic things in this section.

Today we will discuss a few more multiple choice practice questions on geometric optics (geometrical optics):

(1) A thin converging lens produces an inverted real image of an object on a screen. The size of the image is the same as that of the object. When the upper half of the lens is covered by an opaque sheet of paper,

(a) the image disappears

(b) the lower half of the image disappears

(c) the upper half of the image disappears

(d) full image with reduced intensity is obtained

(e) full image having half the size of the object is obtained

Even a small portion of the lens can produce the full image of the object. But the brightness (intensity) of the image will be reduced since half the aperture of the lens admits light. The correct option is (d).

(2) A lens behaves as a converging lens in air and in a liquid of refractive index n₁. But it behaves as a diverging lens in liquids of refractive indices n₂ and n₃. The refractive index of the material of the lens is

(a) greater than both n₂ and n₃

(b) less than n₁ but greater than n₂

(c) less than n₃ but greater than n₂

(d) greater than n₁ but less than both n₂ and n₃

(e) less than n₁ but greater than both n₂ and n₃

The refractive index of the material of the lens must be less than both n₂ and n₃ for the convergent lens to become divergent. Since the lens continues to be convergent in the medium of refractive index n₁, the refractive index of the material of the lens must be.greater than n₁. This follows from lens maker’s equation,

1/f = [(n’/n) – 1] (1/R₁ – 1/R₂) where ‘f’ is the focal length of the lens, n’ is the refractive index of the material of the lens, n is the refractive index of the medium in which the lens is placed and R₁ and R₂ are the radii of curvature of its surfaces. The quantity (1/R₁ – 1/R₂) is positive in the case of a lens that is converging in air. The ratio n’/n has to be greater than 1 for the focal length to be positive (and hence the lens to be converging) when the lens is placed in any other medium.

The correct option is (d).

[A simple ray diagram will lead to the answer to the above question. The adjoining figure shows the manner in which a lens which is converging in air becomes diverging in a medium of refractive index greater than that of the lens. At the first surface of the lens, the ray is refracted away from the normal where as at the second surface the ray is refracted towards the normal. The net effect is to bend the ray further away from the principal axis of the lens].

(3) A thin monochromatic beam of light proceeding through glass is incident at an angle i at glass-air interface. The beam is partially reflected and partially refracted at the interface such that the angle between the reflected and refracted beams is 90º, as shown in the figure. If the angles of reflection and refraction are r and r’ respectively, what is the critical angle for the glass-air interface?.

(a) sin^–1 (r/r’)

(b) sin^–1(r–r’)

(c) sin^–1(tan r)

(d) sin^–1(cos r)

(e) sin^–1(tan r’)

The critical angle C and refractive index n of glass with respect to air is given by

n = 1/sin C ………….(i)

here n = sin r’/sin i = sin r’/sin r since i = r

Since it is given that the angle between the reflected and refracted rays is 90º, we have (with reference to the figure) r’ = (90º – r)

Therefore, n = sin r’/sin r = sin(90º – r)/sin r =cos r/sin r = 1/tan r

Substituting in Eq (i), we have

1/tan r = 1/sinC

Or, sin C = tan r so that C = sin^–1(tan r)

(4) A transparent slab with parallel faces has its refractive index increasing linearly with distance from end A to end B (Fig.). Parallel rays of light are incident normally on this slab as shown. What will happen to these rays? (Ignore reflection at the face of the slab).

(a) They will become convergent after refraction

(b) They will become divergent after refraction

(c) They will bend towards A consequent on refraction

(d) They will bend towards B consequent on refraction

(e) They will proceed undeviated.

Since the angle of incidence is zero (normal incidence), the angle of refraction also is zero irrespective of the refractive index. Therefore the rays will proceed undeviated.

(5) The figure shows five lenses a, b, c, d, and e made of glass. When placed in air which lens/lenses will act as converging lens/lenses?

(a) None

(b) a, c and d

(c) d only

(d) a and d

(e) c and d

Lens (d) is a planoconvex lens and is converging. Lens (c) has one surface convex and the other concave. But the convex surface has smaller radius of curvature. Therefore, the divergence produced because of the concave surface is more than compensated by the convergence produced by the convex surface. It can therefore function as a converging lens.

Lens (a) also has convex and concave faces. But the radius of curvature of the concave face is smaller so that the convergence produced by the convex face is more than compensated by the divergence produced by the concave face. It will act as a diverging lens.

Lenses (b) and (e) are evidently diverging lenses.

So (c) and (d) are the converging lenses [Option (e)].

Geometric Optics for AP Physics B – Additional Multiple Choice Questions

Non-violence leads to the highest ethics, which is the goal of all evolution. Until we stop harming all other living beings, we are still savages.

– Thomas A. Edison

Equations to be remembered in geometric optics were discussed in the post dated 29^th December 2007, followed by some questions with solution in subsequent posts. You can access all those posts either by clicking on the label ‘geometric optics’ (or ‘optics’) below this post or by performing a search within this blog. Today we will discuss a few more multiple choice questions (for practice) in this section:

(1) A converging lens is made of 5 layers of transparent material (fig.). Layers 1, 3 and 5 are of refractive index n₁ where as layers 2 and 4 are of refractive index n₂. A beam of monochromatic light from a distant object, proceeding parallel to the principal axis, is incident on this lens. How many images will be formed?

(a) 1

(b) 2

(c) 3

(d) 5

(e) Nil

Each layer can function as a converging lens of its own focal length f determined by the refractive index n and the radii of curvature R₁ and R₂, as given by lens maker’s equation,

1/f = (n – 1) (1/R₁ – 1/R₂)

Since the lens is made of layers of two different materials, n has two values (n₁ and n₂) and so the focal length f has two values. Therefore there will be two images [Option (b)].

(2) A converging lens is made of 3 portions (1, 2 and 3 in fig.) of three different transparent materials of refractive indices n₁, n₂ and n₃ respectively. A beam of monochromatic light from a distant object, proceeding parallel to the principal axis, is incident on this lens. How many images will be formed?

(a) 1

(b) 2

(c) 3

(d) 6

(e) 9

This lens can be imagined to be made of three different lenses in contact. The combination can have a single focal length and hence there will be just one image [Option (a)].

(3) light is incident normally on a crown glass plate of thickness 3 mm. If the refractive index of crown glass is 1.5 and the speed of light in free space is 3×10⁸ ms^–1 , the time taken by light to travel this thickness of glass is

(a) 1.5×10^–5 s

(b) 3×10^–8 s

(c) 2×10^–11 s

(d) 1.5×10^–11 s

(e) 1.5×10^–12 s

The speed v of light in glass is given by

v= c/n

where c is the speed in free space and n is the refractive index of glass.

Therefore v = 3×10⁸/1.5 = 2×10⁸ ms^–1

The time t taken by light to travel 3 mm (= 3×10^–3 m) thickness of glass is given by

t = 3×10^–3/ 2×10⁸ = 1.5×10^–11 s.

(4) A point source of light (S) is arranged at a depth d in a large water tank. If the critical angle of water is C, what is the area of the water surface that transmits the light from the source?

(a) π d² sin²C

(b) π d² cot²C

(c) π d² tan²C

(d) π d² cos²C

(e) π d² cosec²C

All the light rays within a cone of semi vertical angle C will be transmitted (fig.). The area of the water surface that transmits the light from the source is the area of the circle of radius r given by r = d tanC, as shown in the figure.

[Note that the rays incident at the water surface at an angle C will graze the surface where as the rays incident at at an angle greater than C will be totally reflected back into the water].

Therefore, the area of the water surface that transmits the light from the source is given by

π r² = π (d tanC)² = π d² tan²C

(5) Pick out the incorrect statement:

(a) The refractive index of a material depends on the frequency of the light used for the measurement

(b) The speed of light in a medium depends on the refractive index of the medium

(c) Critical angle of water is smaller for violet light compared to that for red light

(d) In all material media yellow light travels with the same speed

(e) In free space the speed of light is a constant irrespective of the wave length of light.

The speed v of light in a medium is related to the refractive index of the medium as

v = c/n where c is the speed of light in free space and n is the refractive index of the medium.

Different media have different refractive indices for light of a given colour (as determined by the frequency) and for any given medium, the refractive index is maximum for violet light and minimum for red light. There is nothing special about the yellow light and as with light of any other colour, the speed will be different in different material media.

Therefore, the only incorrect statement in the above question is (d).