Question

An object 5cm tall was placed in front of a spherical mirror at 20 cm distance from the mirror. If virtual image of 10cm tall was formed behind the mirror, find the focal length of mirror and position of the image. Name the type of mirror used .​

Answer 1

Answer:

focal lenght=-4cm

Explanation:

Answer 2

Answer :

The focal length of the mirror is -40cm

The position of image is 40cm distance behind the mirror

The type of the mirror is Concave

Given :

• An object of 5cm tall was placed in front of a spherical mirror at distance 20cm from the mirror
• It forms a virtual image which is 10cm tall behind the mirror

To Find :

• The focal length of the mirror
• The position of the image
• Type of mirror

Formulae to be used :

Mirror formula

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

Formula for magnification of mirror

$\sf \star \: \: m = \dfrac{h_{2}}{h_{1}}= -\dfrac{v}{u}=\dfrac{f}{f-u} =\dfrac{f-v}{f}$

Solution :

Given ,

h₁ = 5cm

h₂ = 10cm

u = -20cm

Applying the formula of magnification

$\sf \implies m = \dfrac{h_{2}}{h_{1}} = \dfrac{f}{f - u }\\\\ \sf \implies \dfrac{10}{5}=\dfrac{f}{f-(-20)} \\\\ \sf \implies 2 = \dfrac{f}{f+20} \\\\ \sf \implies 2f + 40 = f \\\\ \sf \implies 2f - f = -40 \\\\ \sf \implies f = -40cm$

Therefore , the focal length of the mirror is -40cm , the negative sign indicates the type of mirror is concave .(since concave mirror's focus is in front of it )

Now by mirror formula

$\sf \implies \dfrac{1}{v}+\dfrac{1}{-20}=\dfrac{1}{-40} \\\\ \sf \implies \dfrac{1}{v} = \dfrac{1}{20} -\dfrac{1}{40} \\\\ \sf \implies \dfrac{1}{v} = \dfrac{1}{40} \\\\ \sf \implies v = 40cm$

Thus the image is formed 40cm behind the mirror.