Refraction by Spherical Surfaces

Practice Questions:

Question 1: easy

An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance \(x\) in front of the second wall. The required focal length of the lens will be

1. \(\frac{x}{2}\)

2. \(\frac{x}{4}\)

3. Less than \(\frac{x}{4}\)

4. \(\frac{x}{4}\) but less than \(\frac{x}{2}\)

View Answer

For an image of equal size, the magnification is \(m = -1\), which means the image distance \(v = x\) is equal to the object distance \(u = x\). Using the lens formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{2}{x} ⇒ f = \frac{x}{2}\).

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Question 2: easy

A convex lens ‘A’ of focal length 20 cm and a concave lens ‘B’ of focal length 5 cm are kept along the same axis with a distance ‘\(d\)’ between them. If a parallel beam of light falling on ‘A’ leaves ‘B’ as a parallel beam, then the distance ‘\(d\)’ in cm will be

1. 30

2. 25

3. 15

4. 50

View Answer

For an incoming parallel beam to emerge parallel from the combination, the focus of the first lens must coincide with the virtual focus of the second lens. Thus, \(d = f_1 - |f_2| = 20 - 5 = 15\) cm.

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