Question

When the object is at distance $u_1$ and $u_2$ the images formed by the same lens are real and virtual respectively having same magnification. The focal length of the lens is :

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

Answer 1

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

For convex lens

$\frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1}$

$\Rightarrow \frac{u_1}{v_1}+1=\frac{v_1}{F}......\left(1\right)$

For virtual image

$\frac{-1}{v_2}+\frac{1}{u_2}=\frac{1}{f}$

$\frac{-u_2}{v_2}+1=u_2/F.......\left(2\right)$

$m_1=m_2$

$v_1{u}_1=v_2{u}_2$...................(3)

Add $\left(1\right)\left(2\right)$ and use $3$

$f=(u_1+u_2)/2$