Question

When the 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 focal length of the lens is:

• A. $\frac{1}{2}\sqrt{{\mu }_1{\mu }_2}$
• B. $\frac{{\mu }_1+{\mu }_2}{2}$
• C. $\sqrt{{\mu }_1{\mu }_2}$
• D. $\sqrt{({\mu }_1+{\mu }_2)}$

Answer 1

Answer: (B)

$\frac{{\mu }_1+{\mu }_2}{2}$

For real image,

$\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f}$..........(1)

$\Rightarrow \frac{u_1}{v_1}+1=\frac{u_1}{f}$..........(2)

For virtual image,

$-\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f}$.........(3)

$-\frac{u_2}{v_2}+1=\frac{u_2}{f}$...........(4)

Manification, $=\frac{v_1}{u_1}=\frac{v_2}{u_2}$..........(5)

Adding (2) and (4), and substituting the same in (5), we get,

$2=\frac{u_1}{f}+\frac{u_2}{f}$

$\Rightarrow f=\frac{u_1+u_2}{2}$