Refraction of Light at Curved Surfaces — Practice

Question 1

Which one of the following materials cannot be used to make a lens?

Accepted Answer

clay

Question 2

Which of the following is true for a convex lens?

Accepted Answer

the distance of the virtual image is not greater than the object distance

Question 3

The focal length of a plano-convex lens with radius of curvature R and refractive index n is

Accepted Answer

f = R/(n - 1)

Question 4

The focal length of a converging lens is 20 cm. An object 2 cm high is placed 60 cm from the lens. Where is the image formed, what kind of image is it, and what is its height?

Accepted Answer

With f = 20 cm and u = -60 cm, the lens formula gives 1/v = 1/f + 1/u = 1/20 - 1/60 = 2/60, so v = 30 cm. The image forms 30 cm from the lens on the far side and is real. Magnification m = v/u = 30/(-60) = -0.5, so the image height is -0.5 x 2 = -1 cm. The image is real, inverted and diminished, 1 cm high, formed 30 cm from the lens.

Question 5

A double convex lens has two surfaces of equal radii R and refractive index n = 1.5. Find the focal length f.

Accepted Answer

For a double convex lens with equal radii, the sign convention gives R1 = +R and R2 = -R. Substituting into the lens maker's formula, 1/f = (1.5 - 1)(1/R - 1/(-R)) = 0.5 x (2/R) = 1/R. So f = R: the focal length equals the radius of curvature.

Question 6

How do you verify experimentally that the focal length of a convex lens increases when it is kept in water?

Accepted Answer

Note the focal length of the lens in air from the lab activity. Take a vessel about four times that depth, place a black stone at the bottom, and fill with water so the depth above the stone exceeds the focal length in air. Dip the lens horizontally on a circular holder, setting the stone-to-lens distance equal to or less than the air focal length, and look at the stone through the lens. You can still see the image even beyond the air focal length, and only when raised much further does it disappear. Since the image is still visible at a separation greater than the air focal length, the focal length in water must be larger.

Question 7

A parallel beam is incident on a converging lens of focal length 40 cm. Where should a diverging lens of focal length 15 cm be placed so that the beam remains parallel after passing through both lenses?

Accepted Answer

The parallel beam would converge to the focus of the first lens, 40 cm from it. For the emerging beam to be parallel again, that convergence point must coincide with the focus of the diverging lens, so that rays heading towards its focus emerge parallel. The diverging lens has focal length 15 cm, so it must be placed 40 - 15 = 25 cm from the converging lens, with the convergence point 15 cm beyond it.

Question 8

You are inside a swimming pool near the edge and a friend is standing on the edge. Do you see your friend taller or shorter than usual, and why?

Accepted Answer

Taller. Light from your friend travels from air into water, that is from a rarer to a denser medium, and refracts towards the normal as it enters. Working with the plane-surface formula n2/v = n1/u, with n1 = 1 for air and n2 = n for water, the image distance comes out as n times the object distance, and since n is greater than 1 the apparent distance is greater. The parts of your friend farthest from the surface appear displaced upwards more than the nearer parts, so the friend appears stretched and taller than usual.

Question 9

A convex lens is made up of three different materials, arranged in horizontal layers. How many images does it form?

Accepted Answer

Three. Each material has its own refractive index, so by the lens maker's formula each layer has its own focal length even though the two surface radii are shared. Light passing through each layer therefore converges at a different point on the principal axis, giving three separate images of the same object.

Question 10

Two converging lenses are placed in the path of parallel rays so that the rays remain parallel after passing through both. How should they be arranged?

Accepted Answer

They must be separated by the sum of their focal lengths, so that the focal point of the first coincides with the focal point of the second. The parallel beam converges to the focus of the first lens; those rays then arrive at the second lens as if coming from its own focus, and by the fourth ray behaviour they emerge parallel to the principal axis.

Practice — Refraction of Light at Curved Surfaces

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