Question

When the real object is at distances $u_1$ and $u_2$, the images formed by the same lens are real and virtual respectively, and of the same size. Then the focal length of the lens is

• A. $\frac{1}{2}\sqrt{u_1{u}_2}$
• B. $\frac{1}{2}(u_1+u_2)$
• C. $\sqrt{u_1{u}_2}$
• D. $2(u_1+u_2)$

Answer 1

The correct option is B $\frac{1}{2}(u_1+u_2)$

For case $1$ : real image,
$u=-u_1$
$v=K{u}_1$

$\therefore \frac{1}{K{u}_1}+\frac{1}{u_1}=\frac{1}{f}$ .................$\left(i\right)$

For case $2$ : virtual image of same size as real image of case $1$,
$u=-u_2$
$v=-K{u}_2$

$\therefore \frac{1}{-K{u}_2}+\frac{1}{u_2}=\frac{1}{f}$ .................$\left(ii\right)$

On solving $\left(i\right)$ and $\left(ii\right)$, we get,
$f=\frac{1}{2}(u_1+u_2)$