Question

An object of size 4 cm is placed at a distance of 30 cm in front of a concave mirror of radius of curvature 20 cm. Find the position ($v$) and size ($h_i$) of the image.

• A.
  $v$ = 15 cm, $h_i$ = -4 cm
  
• B. $v$ = -15 cm, $h_i$ = -2 cm
• C.
  $v$ = 40 cm, $h_i$ = - 1.5 cm
• D.
  $v$ = 5 cm, $h_i$ = 4 cm
  

Answer 1

The correct option is B $v$ = -15 cm, $h_i$ = -2 cm

Given,
Height of object, $h_o$ = + 4 cm (Object is above principal axis)

Object distance, $u$ = -30 cm (Object is in front of the mirror)
Radius of curvature, R = -20 cm (Concave mirror)

Since,
$f=\frac{R}{2}f=\frac{-20}{2}=-10cm\text{Using Mirror equation}\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\frac{1}{-10}=\frac{1}{-30}+\frac{1}{v}⟹v=-15cm$
(negative sign implies that image is formed in front of the mirror)

Therefore, the image is formed at a distance of 15 cm in front of the mirror.
$\text{Using magnification formula,}m=\frac{-v}{u}=\frac{h_i}{h_o}⟹m=\frac{-v}{u}⟹m=\frac{-(-15)}{-30}\therefore m=-0.5\text{Since,}m=\frac{h_i}{h_o}-0.5=\frac{h_i}{4}\therefore h_i=-2cm$(negative sign implies that an inverted image is formed)

Therefore, a real and inverted image of height 2 cm is formed at 15 cm in front of the mirror.