Light - Reflection and Refraction — Question 14

Given,

Object distance (u) = -20 cm

Object height (h) = 5 cm

Image distance (v) = ?

Image height (h) = ?

Radius of curvature (R) = 30 cm

We know, R = 2f or f = $\dfrac{R}{2}$

Hence, f = $\dfrac{30}{2}$ = 15 cm

According to the mirror formula,

$\dfrac{1}{v}$ + $\dfrac{1}{u}$ = $\dfrac{1}{f}$

Substituting the values we get,

$\dfrac{1}{v} + \dfrac{1}{-20} = \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 20} + \dfrac{1}{15} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{3+4}{60} \\[1em] \Rightarrow \dfrac{1}{v} = \dfrac{7}{60} \\[1em] \Rightarrow v = \dfrac{60}{7} = 8.57\text{ cm}$

Image is virtual, erect and formed 8.57 cm behind the mirror.

$\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \\[0.5em] \dfrac{-8.57}{-20} = \dfrac{\text{height of image}}{5} \\[0.5em] \Rightarrow \text{height of image} = \dfrac{-8.57 \times 5}{-20} \\[0.5em] \Rightarrow \text{height of image} = 2.14 \text{ cm}$

Height of image is 2.14 cm