Question

The image of a stone is focussed on a screen using a concave mirror in air. The image is formed at a distance of $20\text{m}$ when the object is kept at a distance of $10\text{m}$. If the mirror is replaced by another mirror of twice the focal length, determine the distance by which the screen has to be moved so that the image is focussed on it.

• A. $3.3\text{m}$ to the right
• B. $3.3\text{m}$ to the left
• C. $6.6\text{m}$ to the left
• D. Cannot be captured on the screen

Answer 1

The correct option is D Cannot be captured on the screen

Given,
object distance, $u=10\text{m}$
image distance, $v=20\text{m}$

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\frac{1}{-10}+\frac{1}{-20}=\frac{1}{f}$
$\frac{1}{f}=\frac{-3}{20}$
$⟹f=-\frac{20}{3}\text{m}$

now, since the mirror is replaced by a mirror of twice the focal length, $f^{{}^{\prime }}=2\times f=\frac{40}{3}$
now , new image distance, $v^{{}^{\prime }}=\frac{u\times f^{{}^{\prime }}}{u-f^{{}^{\prime }}}$
$v^{{}^{\prime }}=\frac{-10\times -\frac{40}{3}}{-10-(-\frac{40}{3})}$
$v^{{}^{\prime }}=\frac{40}{3}\text{m}$
the new image distance is $\frac{40}{3}\text{m}$ implies the image is virtual and virtual images cannot be captured on the screen.