Question

Find the focal length of a concave mirror that produces four times larger real image of an object held at 5 cm from the mirror.​

Answer 1

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$

Answer in Attatched File.....

Answer 2

Answer:

Given:

Image size = 4 times (object size)Object held at 5 cm from mirror

To find:

Focal length of the Concave Mirror

Calculation:

Let image distance be v , object distance be u , focal length be f , object height be h_{o} , image height be h_{i}As per the question

:$\boxed{ \bold{ \dfrac{h_{i}}{h_{o}} = - \dfrac{v}{u} = magnification}}$

$\sf{ = > 4 = - \dfrac{v}{u} }$

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

$\sf{= > v = - 4( - 5)}$

$\sf{= > v = 20 \: cm}$

Now applying Mirror Formula :

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

$\: \: \: \sf{ = > \dfrac{1}{f} = \dfrac{1}{20} + \dfrac{1}{( - 5)} }$

$\: \: \: \sf{ = > \dfrac{1}{f} = \dfrac{1}{20} - \dfrac{1}{5} }$

$\: \: \: \sf{ = > \dfrac{1}{f} = \dfrac{(1 - 4)}{20} }$

$\: \: \: \sf{ = > \dfrac{1}{f} = \dfrac{( - 3)}{20} }$

$\: \: \: \sf{ = > f = - \dfrac{20}{3} }$

$\: \: \: \sf{ = > f = - 6.67 \: cm}$

So final answer :

$\boxed{ \red{ \huge{ \bold{\sf{ f = - 6.67 \: cm}}}}}$

Negative sign verifies that the mirror is concave in nature.

Explanation: