Question

A thin double convex lens of focal length f is broken into two equal halves at the axis. The two halves are combined as shown in figure. What is the focal length of combination in (ii) and (iii).

Answer 1

Answer:

Part a)

For first combination we have

$f_{eq} = infinite$

Part b)

For second combination of lens we have

$f_{eq} = \frac{f}{2}$

Explanation:

Part a)

As we know that focal length of combination of lens is given as

$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$

now we know that both parts have same medium and same radius of curvature

so we have

$f_1 = f_2 = f$

in figure (ii) both parts are inverted with respect to each other

so we have

$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{-f}$

$\frac{1}{f_{eq}} = 0$

$f_{eq} = infinite$

Part b)

Again for the combination of lens we will use

$\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$

now we know that both parts have same medium and same radius of curvature

so we have

$f_1 = f_2 = f$

in figure (iii) both parts are are in same orientation so we have

$\frac{1}{f_{eq}} = \frac{1}{f} + \frac{1}{f}$

$\frac{1}{f_{eq}} = \frac{2}{f}$

$f_{eq} = \frac{f}{2}$