Question

A small candle, $2.5\, cm$ in size is placed at $27\, cm$ in front of a concave mirror of radius of curvature $36\, cm$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

• A. 54 cm
• B. 27 cm
• C. 28 cm
• D. 475 cm

Answer 1

Answer: (A)

Given, radius of curvature of concave mirror, $R=-36\, m$
(For concave mirror radius of curvature is taken as negative)
$\therefore $ Focal length $f=\frac{R}{2}=\frac{36}{2}=-18 \,cm$
Distance of object $u=-27\, cm$
(Object distance is always taken as negative)
Height of object $ O=2.5 \,cm$
Use the mirror formula
$\frac{1}{f}=\frac{1}{v} +\frac{1}{u}$
$\Rightarrow -\frac{1}{-18}=\frac{1}{-27}-\frac{1}{ 54} $
$\frac{1}{v}=-\frac{1}{18}+\frac{1}{27}=\frac{-3+2}{54}=-\frac{1}{54}$
Distance of screen from mirror $v=-54 \,cm$
Let the size of image be $I$.
By using the formula of magnification for mirror
$m =-\frac{v}{u}=\frac{I}{O} $
$\frac{-(-54)}{-27} =\frac{1}{2.5} $
$I =-5\,cm$
The negative sign shows that the image is formed in front of the mirror and it is inverted.
Thus, the screen should be placed at a distance $54\, cm$ and the size of image is $5\, cm$, real, inverted and magnified in nature.