Question

When an object is placed at a distance of $25cm$ from a mirror, the magnification is $m_1$. The object is moved $15cm$ away with respect to the earlier position, magnification becomes $m_2$, If $\frac{m_1}{m_2}=4$, the focal length of the mirror is

Answer 1

$u_1=-25cm,u_2=-(25+15)=-40cm$

$m_1=-\frac{v_1}{u_1}$

$m_2=-\frac{v_2}{u_2}$

$\frac{m_1}{m_2}=\frac{v_1}{v_2}\times \frac{u_2}{u_1}$

$4=\frac{v_1}{v_2}.\frac{40}{25}$

$v_1=2.5v_2$

$\frac{1}{u_1}+\frac{1}{v_1}=\frac{1}{f}\Rightarrow \frac{1}{-25}+\frac{1}{2.5v_2}=\frac{1}{f}$....(1)

$\frac{1}{u_2}+\frac{1}{v_2}=\frac{1}{f}\Rightarrow \frac{1}{-40}+\frac{1}{v_2}=\frac{1}{f}$....(2)

Divinding (2) by 2.5 , $\frac{1}{-100}+\frac{1}{2.5v_2}=\frac{1}{f}$......(3)

$\left(1\right)-\left(3\right)\Rightarrow \frac{1}{-25}+\frac{1}{100}=\frac{1}{f}-\frac{1}{2.5f}$

$\frac{-3}{100}=\frac{1.5}{2.5f}$

$f=-20cm$