Question

When an object is placed at a distance of $25\text{cm}$ from a mirror, the magnification is $m_1$. But the magnification becomes $m_2$ when the object is moved $15\text{cm}$ farther away from the mirror with respect to earlier position. If $\frac{m_1}{m_2}=4$, then find the focal length of the mirror and its nature.

• A. $20\text{cm}$, convex
• B. $20\text{cm}$, concave
• C. $10\text{cm}$, convex
• D. $10\text{cm}$, concave

Answer 1

The correct option is B $20\text{cm}$, concave

Given:
$\frac{m_1}{m_2}=4$
$\Rightarrow \frac{f-u_2}{f-u_1}=4$ $[\because m=\frac{f}{f-u}]$
$\Rightarrow \frac{f-(-40)}{f-(-25)}=4$
On solving this, we get,
$f=-20\text{cm}$
As focal length is negative, the mirror is concave.
$\begin{array}{c}\text{Why this question?}\\ \text{Image is magnified when object is placed between C and F and diminished when object is placed beyond C.}\\ \end{array}$