AP Physics B - Multiple Choice Practice Questions on Geometric Optics

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– Mark Twain

Questions on geometric optics were discussed on many occasions on this site. Try a search for ‘geometric optics’ using the search box provided on the side bar this page or click on the label ‘geometric optics’ below this post, to access all posts related to geometric optics.

Today we shall discuss a few more multiple choice practice questions on geometric optics:

(1) Focal power or optical power (or simply, power) of a lens is the reciprocal of its focal length. The focal length has to be expressed in metre for obtaining the power in its popular unit, dioptre.

Two thin converging lenses of focal lengths 40 cm and 60 cm are kept in contact so as to have a common principal axis. The power of this combination is approximately

(a) 1

(b) 2

(c) 3

(d) 4

(e) 5

If the focal length of the combination of the lenses is F metre, the power of the combination is 1/F.

When two thin lenses of focal lengths f₁ and f₂ are kept in contact so as to have a common principal axis, the focal length F of the combination is given by the reciprocal relation,

             1/F = 1/f₁ + 1/f₂

[In fact, this is a relation connecting the powers of the individual lenses to the power of the combination]

Therefore we have 1/F = 1/0.4 + 1/0.6 = (0.6 + 0.4)/(0.4×0.6) = 1/0.24 ≈ 4

(2) A bright object O is placed on the principal axis of a thin converging lens. The lens  produces a real magnified image of O at the position I (Fig.) at a distance of 44 cm from the lens. When a convex mirror of focal length 12 cm is interposed between the image and the lens as shown in the figure, a real image of the same size as the object is obtained side by side with the object. What is the distance between the lens and the mirror?

(a) 12 cm

(b) 16 cm

(c) 20 cm

(d) 22 cm

(e) 24 cm

Since the final image is obtained side by side with the object, the rays of light must fall normally on the mirror. This means that the initial image I must be formed at the centre of curvature of the mirror. The radius of curvature of the mirror is twice its focal length and is therefore equal to 24 cm. The initial image I must therefore be at a distance of 24 cm from the mirror. Since the distance between the lens and the initial image I is 44 cm, the distance between the lens and the mirror must be (44 – 24) cm = 20 cm.

(3) An object is placed in front of a convex mirror of focal length f . The distance of the object is greater than f  but less than 2f. The image of the object is

(a) in front of the mirror at distance less than f

(b) in front of the mirror at distance greater than f  but less than 2f

(c) in front of the mirror at distance 2f

(d) behind the mirror at distance less than f

(e) behind the mirror at distance greater than f  but less than 2f

You may try drawing a ray diagram to locate the image, as shown in the adjoining figure. A ray of light proceeding towards the centre of curvature C of the mirror gets reflected from the mirror and retraces its path. Another ray proceeding parallel to the principal axis gets reflected from the mirror and proceeds as shown, as though it diverges away from the focus F of the mirror. The image is virtual and is located behind the mirror at distance less than f.

The correct option is (d).

[Note that the image will be located behind the mirror at distance less than f for all object distances].

(4) A student tries to burn a sheet of paper by focusing sun light on it using a converging lens of focal length f. The radius of the sun is R and its distance from the earth is D. The diameter of the image of the sun obtained on the sheet of paper is

(a) 2Rf/D 

(b) 2RD/f 

(c) RD/f 

(d) RD/2f 

(e) Rf/2D 

Since the sun is far away from the lens, a real diminished image of the sun is formed at the focus of the lens.

[The student has to keep the sheet of paper at the focal plane of the lens for burning it].

The formation of the image is shown in the adjoining figure in which the radius of the image is shown as r.

From the similar triangles, we have

             R/r = D/f

Or, r = Rf/D

The diameter of the image = 2r = 2Rf/D  

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).