Question

An object placed at 20cm distance from concave mirror for obtaining 4 times of
image of the object height. What will be focal length of mirror.
​

Answer 1

Given:

An object placed at 20cm distance from concave mirror for obtaining 4 times of image of the object height.

To find:

Focal length of mirror?

Calculation:

• Image height = 4 times object height. So, we can say:

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

$\implies 4= - \dfrac{v}{u}$

$\implies - 4u= v$

Applying Mirror Formula:

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

$\rm \implies\: \dfrac{1}{f} = \dfrac{1}{ - 4u} + \dfrac{1}{u}$

$\rm \implies\: \dfrac{1}{f} = \dfrac{3}{ 4u}$

$\rm \implies\: \dfrac{1}{f} = \dfrac{3}{ 4 (- 20)}$

$\rm \implies\: f = - 26.67 \: cm$

So, focal length is 26.67 cm.

Answer 2

Given:

An object placed at 20cm distance from concave mirror for obtaining 4 times of image of the object height.

To find:

Focal length of mirror?

Calculation:

• Image height = 4 times object height. So, we can say:

$\frac{hi}{ho} = - \frac{v}{u}$

$= > 4 = - \frac{v}{u}$

$= > - 4u = v$

$= > - 4 u= v$

Applying Mirror Formula:

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

$= > \frac{1}{f } = \frac{1 }{ - 4u} + \frac{1}{u}$

$= > \frac{1}{f} = \frac{3}{4u}$

$= > \frac{1}{f} = \frac{3}{4( - 20)}$

$= > f = - 26.67$

So, focal length is 26.67 cm.