Question

When the object is at distance $u_1$ 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. $\frac{u_1+u_2}{2}$
• C. $\sqrt{u_1{u}_2}$
• D. $\sqrt{(u_1+u_2)}$

Answer 1

Answer: (B)

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

$\begin{array}{c}\text{* by apply ing mirror formula in tooth the cases.}\\ \Rightarrow \frac{1}{-V_1}-\frac{1}{(-V_1)}=\frac{1}{f}=\left(1\right)\Rightarrow \frac{1}{(-V_2)}-\frac{1}{(-u_2)}=\frac{1}{f}=\left(2\right)\end{array}$

by applying magnification formula $\Rightarrow m=\frac{-v_1}{M_2}-\left(5\right)m=\frac{V_2}{u_2}$

as maquification is same in both cases by sign of

V. \& U , are different in $1^{\text{st}}$ case.

$\begin{array}{c}\text{by eq (1) we can say}\Rightarrow v_1=\frac{u_{1}f}{u_1-f};\left(5\right)=v_2=\frac{u_{2}f}{f-u_2}=\left(6\right)\\ \text{by (3)}\&(4\Rightarrow \frac{v_1}{{\mu }_1}=\frac{v_2}{u_2}{-}^{}\text{(7)}\end{array}$

$\begin{array}{c}\text{by equ (5), (6)(7) as}\frac{v_1}{x_2}=\frac{u_1}{u_2}\\ \Rightarrow \frac{\left(\frac{u_{1}f}{u_1-f}\right)}{\frac{u_{2}f}{f-u_2}}=\frac{u_1}{u_2}\end{array}$

$\frac{\frac{f}{u_1-f}}{\frac{f}{f-u_2}}=1$

$f-u_2=u_1-f$

$2f=u_1+u_2$

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

Hence obtion $B$ 's correct