Two converging lens have focal length $f = 3$ cm and $f = 5$ cm. The optical axes of the lenses coincide. This lens system is used to form an image of an object kept at a certain distance from the lens. In this situation, the size of the image does not depend on the distance of the object from the lens. What is the distance (in cm) between the two lens.

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Solution

Given: $f_{1} = 3 c m$ and $f_{2} = 5 c m$
To find : d
Concept : $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
v is image distance. u is object distance. f is focal length.
Solution : Let initial position of object be u and position of image by lens 1 be $v_{1}$ and distance of final image be $v$ .
So for lens 1: $\frac{1}{v_{1}} - \frac{1}{- u} = \frac{1}{f}$
$\frac{1}{v_{1}} + \frac{1}{u} = \frac{1}{3}$ $\to (1)$
and for lens 2
$\frac{1}{v} + \frac{1}{d - v_{1}} = \frac{1}{5}$ $\to (2)$
from (1) we get $v_{1} = \frac{1}{\frac{1}{3} - \frac{1}{4}}$
from (2) we get $v_{1} = d - \frac{1}{\frac{1}{5} - \frac{1}{v}}$
So by equating $v_{1}$ from (1) and (2) we get
$\frac{1}{\frac{1}{3} - \frac{1}{u}} = d - \frac{1}{\frac{1}{5} - \frac{1}{v}}$
$\frac{3 u}{u - 3} = d - \frac{5 v}{v - 5}$
$\frac{3 u}{u - 3} = \frac{d v - 5 d - 5 v}{v - 5}$
$3 u v - 15 u = d u v - 5 d u - 5 u v - 3 d v + 15 d + 15 v$
$3 d v - 15 d - 15 v = (d - 8) u v + (15 - 5 d)$
$(3 d - 15) v - (d - 8) u v = (15 - 5 d) u + 15 d$
$v = \frac{(15 - 5 d) u + 15 d}{(3 d - 15) - (d - - 8)}$
Hence for v to be independent of u $15 - 5 d = 0$
so $d = 3$

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