Reflection of Light — Question 10

Given,

(a) Distance of the object (u) = 30 cm (negative)

Image height = 3 times the height of object

So, magnification (m) = 3 (negative for the real image)

$\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] - 3 = - \dfrac{\text v}{- 30} \\[0.5em] \Rightarrow \text v = 3 \times (- 30) \\[0.5em] \Rightarrow \text v = - 90 \text{ cm} \\[0.5em]$

Hence, the image is formed 90 cm in front of the mirror.

Mirror formula:

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

Substituting the values in the mirror formula we get,

$- \dfrac{1}{30} - \dfrac{1}{90} = \dfrac{1}{\text f} \\[0.5em] \dfrac{1}{\text f} = \dfrac{ - 3 - 1 }{90} \\[0.5em] \dfrac{1}{\text f} = -\dfrac{4}{90} \\[0.5em] \Rightarrow \text f = -22.5 \text{ cm} \\[0.5em]$

Hence, focal length of the mirror = 22.5 cm

(b) The image is formed 90 cm in front of the mirror