Question

The magnification produced by a convex lens for two different positions of an object are $m_1$ and $m_2$ respectively $(m_1>m_2)$. If 'd ' is the separation between the two positions of the object then the focal length of the lens is

• A. $\sqrt{m_1{m}_2}$
• B. $\frac{1}{\sqrt{m_1-m_2}}$
• C. $\frac{d{m}_1{m}_2}{m_1-m_2}$
• D. $\frac{d}{m_1-m_2}$

Answer 1

The correct option is C $\frac{d{m}_1{m}_2}{m_1-m_2}$

$\begin{array}{c}\text{for convex lens}m=\frac{f}{f-u}\\ \text{where}m=\text{magnification}\\ \begin{array}{cc}f& =\text{focal length}\\ u& =\text{object distance.}\end{array}\end{array}$

$m_2=\frac{f}{f-x}-\left(1\right)$

$m_1=\frac{f}{f-(x+d)}$

$f-x-d=\frac{f}{m_1}$

$f-d-\frac{f}{m_1}=x-\left(2\right)$

substitute (2) in (1)

$m_2=\frac{f}{f-(f-d-\frac{f}{m})}$

$m_2=\frac{f}{d+\frac{f}{m_1}}$

$m_{2}d+\frac{m_2}{m_1}f=f$

$m_{2}d=f[1-\frac{m_2}{m}]$

$m_{2}d=f\left(\frac{m_1-m_2}{m_1}\right)$

$f_i=\frac{m_1{m}_{2}d}{m_1-m_2}$

Hence option c is correct