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 :

Let the focal length of the convex lens A be

$f\_1\; =\; +20\backslash \; \backslash text\{cm\}$

and the focal length of the concave lens B be

$f\_2\; =\; -5\backslash \; \backslash text\{cm\}$

(The sign of a concave lens is negative in the thin‑lens sign convention.)

When two thin lenses are separated by a distance $d$, the combined (effective) focal length $F$ is given by the lens‑maker formula

$\backslash frac\{1\}\{F\}=\; \backslash frac\{1\}\{f\_1\}+\; \backslash frac\{1\}\{f\_2\}-\backslash frac\{d\}\{f\_1\; f\_2\}$

In the problem a parallel beam falls on lens A and leaves lens B again as a parallel beam. A system that converts a parallel incident ray into a parallel emerging ray is called *afocal*; its effective focal length is infinite, i.e. $\backslash frac\{1\}\{F\}=0$.

Setting $\backslash frac\{1\}\{F\}=0$ in the formula gives

$0\; =\; \backslash frac\{1\}\{f\_1\}+\; \backslash frac\{1\}\{f\_2\}-\backslash frac\{d\}\{f\_1\; f\_2\}$

Solving for $d$:

$\backslash frac\{d\}\{f\_1\; f\_2\}=\; \backslash frac\{1\}\{f\_1\}+\; \backslash frac\{1\}\{f\_2\}$

$d\; =\; f\_1\; f\_2\backslash !\backslash left(\backslash frac\{1\}\{f\_1\}+\; \backslash frac\{1\}\{f\_2\}\backslash right)$

Factorising the right‑hand side:

$d\; =\; f\_1\; +\; f\_2$

Now substitute the numerical values:

$d\; =\; 20\backslash \; \backslash text\{cm\}\; +\; (-5\backslash \; \backslash text\{cm\})\; =\; 15\backslash \; \backslash text\{cm\}$

Therefore, the required separation between the lenses is

$15\backslash \; \backslash text\{cm\}$