Question

When an object is placed at a distance of $25 \, cm$ from a mirror, the magnification is $m_{1}$ . But the magnification becomes $m_{2}$ , when the object is moved $15 \, cm$ farther away with respect to the earlier position. $\frac{m_{1}}{m_{2}}=4\text{,}$ then find the focal length of the mirror and what type of mirror it is?

• A. $20cm$ , convex
• B. $20cm$ , concave
• C. $10cm$ , convex
• D. $10cm$ , concave

Answer 1

Answer: (B)

$\textit{m}=-\frac{\textit{v}}{\textit{u}}=\frac{\textit{f}}{\textit{f} - \textit{u}}$
so, $\quad m_1=\frac{f}{f-(-25)}=\frac{f}{f+25}$
$\text{Similarly,}\textit{m}_{2}=\frac{\textit{f}}{\textit{f} + \text{40}}$
$\text{and}\frac{\textit{m}_{1}}{\textit{m}_{2}}=4$
$\therefore \frac{\textit{f} + \text{40}}{\textit{f} + \text{25}}=4$
$f=-20cm$
negative sign shows that the mirror is concave.