Question

When an object is at a distance $u_{1}and{u}_2$ from the optical centre of a lens, a real and virtual image are formed respectively, with the same magnification. The focal length of lens is :

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

Answer 1

Answer: (C)

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

$\begin{array}{c}\text{For first case, we use}\\ \frac{1}{v_1}+\frac{1}{u_1}=\frac{1}{f}\\ \text{and for second case, we use}\\ \begin{array}{c}-\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f}\\ \frac{v_2}{4_2}=\frac{f}{f-u_2}\\ \text{But,}\frac{v_1}{u_i}=\frac{v_2}{4_2}\end{array}\end{array}$

$\begin{array}{cc}\therefore \frac{f}{{\mu }_1-f}& =\frac{f}{f-u_2}\\ \therefore f^2-f{u}_2& =f{u}_1-f^2\\ \therefore 2f^2=& f(u_1+u_2)\\ & ;f=\frac{{\mu }_1+{\mu }_2}{2}\\ \text{Hence,}\left(c\right)\text{is correct option.}& \end{array}$