Question

An object is placed in front of a concave mirror focal length $f$. An erect image is formed with a magnification of $2$. To obtain real image with the same magnification, the object has to be moved through a distance of:

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

Answer 1

Answer: (B)

$f$

From mirror formula:

$u=\frac{fv}{v-f}$ ...............eq1,

when the image formed is erect , then linear magnification will be ,

$m=-\frac{v}{u}=2\left(given\right)$ ,

or $v=-2u$ ,

by eq1 ,

$u=\frac{f(-2u)}{(-2u)-f}$ ,

or $u=f/2$ ,

Now, when image formed is real , then linear magnification will be ,

$m^{\prime }=-\frac{v^{\prime }}{u^{\prime }}=-2\left(given\right)$ ,

or $v^{\prime }=2u^{\prime }$ ,

by eq1 ,

$u^{\prime }=\frac{f\left(2u^{\prime }\right)}{\left(2u^{\prime }\right)-f}$ ,

or $u^{\prime }=3f/2$ ,

therefore, movement of the object will be $=u^{\prime }-u=3f/2-f/2=f$ ,