Question

An object is kept on the principal axis of a concave mirror of focal length 12 CM if the object at a distance of 18cm from the mirror Calculate image distance and magnification of the mirror

Answer 1

Given

• Focal length (f) = - 12cm
• Object distance (u) = - 18cm

Find out

• Image distance
• Magnification

Solution

✞ According to the Mirror Formula

$\implies\tt \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f} \\ \\ \\ \implies\tt \dfrac{-1}{18}+\dfrac{1}{v}=\dfrac{-1}{12} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{1}{18}-\dfrac{1}{12} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{2-3}{36} \\ \\ \\ \implies\tt \dfrac{1}{v}=\dfrac{-1}{36} \\ \\ \\ \implies\sf v=-36cm$

Hence, image distance is - 36 cm

$\rule{200}3$

Magnification : It is the ratio of height or size of image to the height or size of object.

$\implies\tt m=\dfrac{-v}{u}=\dfrac{hi}{ho} \\ \\ \\ \implies\tt m=\dfrac{-v}{u} \\ \\ \\ \implies\tt m=\dfrac{-(-36)}{(-18)} = -2$

Hence, magnification is - 2

$\rule{200}3$

Answer 2

Given :

• Focal length (f) = -12cm

• Object distance (u) = -18cm

TO FIND :

• Image distance

• Magnification

SOLUTION :

Applying mirror formula,

$=> \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ => \frac{ - 1}{18} + \frac{1}{u} = \frac{ - 1}{12} \\ \\ => \frac{1}{u} = \frac{1}{18} - \frac{1}{12} \\ \\ => \frac{1}{v} = \frac{12 - 18}{18 \times 12} \\ \\ => \frac{1}{v} = \frac{ - 6}{216} \\ \\ =>\bold\red{ <strong>v</strong> \: <strong>=</strong> <strong>-</strong> <strong>36cm</strong> }\:$

We know that,

$=> Magnification \: = \frac{v}{u} = \frac{h_i}{h_o} \\ \\ => Magnification \: = \frac{ - v}{u} \\ \\ => Magnification \: = \frac{ - ( - 36)}{ - 18} \\ \\ =>\bold\red<strong> </strong>{<strong>Magnification</strong> <strong>=</strong> <strong>-</strong> <strong>2</strong><strong> </strong>}$

Hence,

=> Image distance = -36cm

=> Magnification = -2