Question

An object is placed infront of a concave mirror of focal length f. A virtual image is formed with a magnification, the object has to moved by a distance

• A. $f$
• B. $\frac{f}{2}$
• C. $\frac{3f}{2}$
• D. $\frac{2f}{3}$

Answer 1

The correct option is A $f$

In the first case. Let x be the distance of object from the mirror. Then
u = -x
v = + 2x
f = -f
Using $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
or $\frac{1}{2x}-\frac{1}{x}=-\frac{1}{f}$
or $x=\frac{f}{2}$
in the second case, let y be the distance of object from the mirror. Then
u = - y, v = -2y
and f = - f
So $\frac{1}{-2y}-\frac{1}{y}=-\frac{1}{f}$
$\therefore$ $y=\frac{3}{2}f$
So, object will have to be moved by a distance of y - x or f.