Question

A spherical mirror forms an image of magnification $3$ for a real object. If the focal length of the mirror is $24\text{cm}$, then the object distance may be

• A. $32\text{cm}$ only
• B. $16\text{cm}$ only
• C. $32\text{cm}$, $24\text{cm}$
• D. $32\text{cm}$, $16\text{cm}$

Answer 1

The correct option is D $32\text{cm}$, $16\text{cm}$

Given:

Magnification $|m|=3$

Focal length of spherical mirror $f=24\text{cm}$

Since magnification is greater than $1$ for a real object.

The spherical mirror is concave, then we can say that,

Focal length$\left(f\right)=-24\text{cm}$

Magnification from a concave mirror,
$m=\frac{f}{f-u}$

For $m=+3$ , we can write that, $+3=\frac{-24}{-24-u}$

$\Rightarrow u=\frac{-48}{3}=-16\text{cm}$

When object is between pole of concave mirror and focal length, image formed is virtual , erect and magnified.

For $m=-3$ , we can write that , $-3=\frac{-24}{-24-u}$

$\Rightarrow u=\frac{-4\times 24}{3}=-32\text{cm}$

When object is placed between focal length and centre of curvature, then the image formed is real, inverted and magnified.

Hence, we can say that, for a spherical mirror having magnification $3$ , the real object is either at $16\text{cm}\text{or}32\text{cm}$ from the pole of the spherical mirror.

Thus, option (d) is the correct answer.