Question

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

• A. $+30\text{cm}$
• B. $-30\text{cm}$
• C. $+20\text{cm}$
• D. $-20\text{cm}$

Answer 1

The correct option is D $-20\text{cm}$

The magnification of a spherical mirror is -

$m=\frac{f}{f-u}$

Now,

$m_1=\frac{f}{f-(-25)}=\frac{f}{f+25}$

$m_2=\frac{f}{f-[-(25+15\left)\right]}=\frac{f}{f+40}$

According to the question,

$\frac{m_1}{m_2}=4$

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

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

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

Hence, option $\left(D\right)$ is the correct answer.