Question

A convex mirror of focal length $f$ forms an image whose size is $\frac{1}{n}$ times that of the real object. The distance of the object from the mirror is

• A. $(n-1)f$
• B. $\left[\frac{n-1}{n}\right]f$
• C. $\left[\frac{n+1}{n}\right]f$
• D. $(n+1)f$

Answer 1

The correct option is A $(n-1)f$

Given that image size is $\frac{1}{n}$ times that of the real object
A convex mirror forms a virtual and erect image of real object
Magnification $m=+\frac{1}{n}=-\frac{v}{u}$
$\therefore v=-\frac{u}{n}$
From mirror formula,
$\frac{1}{f}=\frac{1}{\left(\frac{-u}{n}\right)}+\frac{1}{u}$
$\frac{1}{f}=\frac{-n}{u}+\frac{1}{u}$
$\frac{1}{f}=\frac{1}{u}(1-n)$

$\therefore u=-(n-1)f$

Hence position of object is at $(n-1)f$ infront of the mirror.

Why this question? Note: Remember that the mirror equation holds for spherical mirrors only at small distances, at larger distances only parabolic mirrors hold this equation