Question

In the displacement method, a convex lens is placed in between an object and a screen. If the magnifications in the two positions are $m_{1}$ and $m_{2}$ and the displacement of the lens between the two positions is $x$, then the focal length of the lens is

• A. $\frac{x}{m_{1}+m_{2}}$
• B. $\frac{x}{m_{1}-m_{2}}$
• C. $\frac{x}{\left(m_{1}+m_{2}\right)^{2}}$
• D. $\frac{x}{\left(m_{1}-m_{2}\right)^{2}}$

Answer 1

Answer: (B)

By principle of reversibility where must be symmetry in the two position

$d=u+v$
$a=v-u$
Hence, $u=\frac{d-a}{2}$ and $v=\frac{d+a}{2}$
Putting in lens equation $\frac{2}{d-a}+\frac{2}{d+a}=\frac{1}{f}$
$\therefore f=\frac{d^{2}-a^{2}}{4 d}$
$m_{1}=\frac{d+a}{d-a}$
$m_{2}=\frac{d-a}{d+a}$
$\left|m_{1}-m_{2}\right|=\frac{4 d a}{d^{2}+a^{2}}$
$\left|m_{1}-m_{2}\right|=\frac{a}{f} .(a=x)$
$f=\frac{x}{\left|m_{1}-m_{2}\right|}$