Question

A 2cm high object is placed at a distance of 20cm from a concave mirror. A real image is formed at 40cm from the mirrpr.Calculate the focal length of the mirror.​

Answer 1

Answer:-

f = -13. 33cm

Given:-

Height of object = 2cm

Object distance (u) = -20 cm

Image distance (v) = -40 cm

To find :-

Focal length of mirror.

Solution:-

From Mirror formula we have,

★ $\boxed{\sf{ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}$

Now, put the given value,

→$\dfrac{1}{f}=\dfrac{-1}{40}+\dfrac{-1}{20}$

→$\dfrac{1}{f}= \dfrac{-1}{40}-\dfrac{1}{20}$

→$\dfrac{1}{f}= \dfrac{-1-2}{40}$

→$\dfrac{1}{f}= \dfrac{-3}{40}$

→$f = \dfrac{-40}{3}$

→$f = -13. 33cm$

hence,

The focal length of mirror will be -13.33cm.

Answer 2

Answer:

Focal length is $-\dfrac{40}{3} = 13.333cm$

Explanation:

Given Problem:

A 2cm high object is placed at a distance of 20cm from a concave mirror. A real image is formed at 40cm from the mirrpr.Calculate the focal length of the mirror.​

Solution:

To Calculate:

Focal Length.

---------------------

Method:

Given that,

Object distance (u) = -20cm.

Image distance (v) = -40cm.

Height of object $(h_0)$ = 2cm.

We know that,

★Mirror Formula:★

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

$\implies\dfrac{1}{-40} +\dfrac{1}{-20} = \dfrac{1}{f}$

$\implies\dfrac{-1-2}{40} = \dfrac{1}{f}$

$\implies\ f = -\dfrac{40}{30} = - 13.333cm$

Hence,

The focal length is $-\dfrac{40}{30} = - 13.333cm$