Question

An object placed in front of a concave mirror at a distance of $xcm$ from the pole gives a $3$ times magnified real image. If it is moved to a distance of $(x+5)cm$, the magnification of the image becomes $2$. The focal length of the mirror is

• A. $15cm$
• B. $20cm$
• C. $25cm$
• D. $30cm$

Answer 1

The correct option is D $30cm$

Magnification $=\frac{-v}{u}$

where, $v=$ image distance and $u=$ object distance.

We then have:

$-v_1=-3\times -x\Rightarrow v_1=-3x-v_2=2(x+5)\Rightarrow v_2=-2x-10$.

By the mirror formula, we have:

$\Rightarrow \frac{1}{u_1}+\frac{1}{v_1}=\frac{1}{f}=\frac{1}{u_2}+\frac{1}{v_2}\Rightarrow \frac{1}{x}+\frac{1}{3x}=\frac{1}{x+5}+\frac{1}{2x+10}\Rightarrow \frac{4}{3x}=\frac{3}{2x+10}\Rightarrow 8x+40=9x\Rightarrow x=40\Rightarrow \frac{1}{f}=\frac{1}{u_1}+\frac{1}{v_1}=\frac{1}{x}+\frac{1}{3x}=\frac{4}{3x}=\frac{1}{30}\Rightarrow f=30cm$

Thus, the focal length of the mirror is 30cm.