Question

The image of an object placed on the principal axis of a concave mirror of focal length 12 cm is formed at a point which is 10 cm more distant from the mirror than the object. The magnification of the image is

• A. 2
• B. 1.5
• C. 1
• D. 2.5

Answer 1

The correct option is B 1.5

Focal length = -12 cm
Let u be the object distance and v be image distance,

then we can write $\frac{1}{u}+\frac{1}{v}=\frac{1}{-12}$ $\cdots \left(i\right)$
Given, v = u + 10

Now putting values
$\therefore \frac{1}{-u}+\frac{1}{-(u+10)}=\frac{1}{-12}$

Or,
$\Rightarrow \frac{u+10+u}{(u+10)u}=\frac{1}{12}$

Or,
$\Rightarrow u^2-14u-120=0$
$\Rightarrow$ u=20

So we can write
$\Rightarrow$ u = 20 cm and v = 30 cm

Now,
magnification $\frac{v}{u}=\frac{30}{20}=\frac{3}{2}$
Hence, m = 1.5