Question

When an object is placed at a distance of $25\text{cm}$ from a mirror, the magnification of the real image formed is $m_1$. When the object is moved $15\text{cm}$ away with respect to the earlier position along the principal axis, magnification becomes $m_2$. If $m_1\times m_2=4$, the focal length of the mirror in $\text{cm}$ is $P\times 10$. Then, find the value of $P$

Answer 1

Initially, let the object distance be $u$.
Then, applying mirror formula,
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
or $v=\frac{fu}{u-f}$
Magnification, $m=\frac{-v}{u}=\frac{f}{f-u}$

Initially, $u=-25\text{cm}$ and after moving, $u^{\prime }=-40\text{cm}$

Given, $m_1\times m_2=4$
$\Rightarrow \frac{f}{f+25}\times \frac{f}{f+40}=4$
$\Rightarrow f^2=4(f+25)(f+40)=4(f^2+40f+25f+1000)$
$\Rightarrow f^2=4[f^2+65f+1000]$
$\Rightarrow 3f^2+260f+4000=0$

$\Rightarrow f=-20\text{cm}$ or $-66.67\text{cm}$
Since image is given to be real, we can neglect $-66.67\text{cm}$
Hence, $P=2$