Question

When an object is placed at a distance of $25\text{cm}$ from a mirror, the magnification is $m_1.$ The object is moved $15\text{cm}$ away with respect to the earlier position along principal axis, magnification becomes $m_2.$ If $\frac{m_1}{m_2}=4.$ The magnitude of focal length of the mirror in $\text{cm}$ is $n\times 10\text{cm}.$ The value of $n$ is (integer only)

• A. 2.00
• B. 2.0
• C. 2

Answer 1

Answer: (A), (B), (C)

Given,
$u_i=-25\text{cm};u_f=-40\text{cm}\frac{m_1}{m_2}=4$

As we know, $m=-\frac{v}{u}=\frac{f}{f-u}$

$\Rightarrow m_1=\frac{f}{f+25}\dots \left(1\right)$

$\Rightarrow m_2=\frac{f}{f+40}\dots \left(2\right)$

From (1) and (2) we get,

$\frac{m_1}{m_2}=\frac{f+40}{f+25}[\because \frac{m_1}{m_2}=4]$

$4=\frac{f+40}{f+25}$

$4f+100=f+40$

$\Rightarrow f=-20\text{cm}$

The negative sign indicates that it is a concave mirror.

$\therefore n=2$