Question

If in the following figure, height of object is $H_{1}=+2.5 \,cm$, then height of image $H_{2}$ formed is Concave mirror

• A. -5 cm
• B. +5 cm
• C. +7.5 cm
• D. -7.5 cm

Answer 1

Answer: (A)

-5 cm

Answer 2

Magnification is ratio of size of image to that of object.
Here, $u=-1.5 \,\,f, f=-f$
From concave mirror formula,
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ we have,
$\frac{1}{(-f)}=\frac{1}{v}+\frac{1}{(-1.5 f)}$
$\therefore \frac{1}{v}=-\frac{1}{v}+\frac{1}{(-1.5 f)}$
$\therefore \frac{1}{v}=-\frac{1}{3 f}$ Or $v=-3 f$
Now, magnification in mirror is given by
$m=\frac{I}{O}=\frac{v}{-u}$
where $I$ is size of image,
and $O$ is size of object.
Here, $0=+2.5 \,cm$
$\therefore \frac{I}{2.5}=\frac{-3 f}{-(-1.5 f)}$
$\therefore I=-5 \,cm$
Note: In case of a concave mirror, magnification may be positive or negative, but in case of convex mirror magnification is positive only.