Question

22. An object of height 1.5 m is placed at a distance of 15 cm from a concave mirror of focal
length 30 cm so that a virtual image is formed. Calculate position nature and size of the
image. Draw a ray diagram to show the image formation in the above case​

Answer 1

Given:

An object of height 1.5 m is placed at a distance of 15 cm from a concave mirror of focal length 30 cm so that a virtual image is formed.

To find:

• Image position
• Image size
• Ray diagram

Calculation:

Applying MIRROR FORMULA:

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

$\implies \dfrac{1}{( - 30)} = \dfrac{1}{v} + \dfrac{1}{( - 15)}$

$\implies - \dfrac{1}{30} = \dfrac{1}{v} - \dfrac{1}{ 15}$

$\implies \dfrac{1}{v} = \dfrac{1}{ 15} - \dfrac{1}{30}$

$\implies \dfrac{1}{v} = \dfrac{2 - 1}{30}$

$\implies \dfrac{1}{v} = \dfrac{1}{30}$

$\implies \: v = 30 \: cm$

So, +ve sign of 'v' shows that virtual image is formed.

$\dfrac{h_{i}}{ h_{o} } = - \dfrac{v}{u}$

$\implies \dfrac{h_{i}}{ h_{o} } = - \dfrac{30}{( - 15)}$

$\implies \dfrac{h_{i}}{1.5} = - \dfrac{30}{( - 15)}$

$\implies \dfrac{h_{i}}{1.5} = 2$

$\implies h_{i} = 3 \: m$

So, image is magnified.