Question

0.3. A spherical mirror produces erect image of height 2.5 cm for an object of height 2.0 cm.(a) Identify the mirror,(b) Find the magnification of the image,(c) Find the exact location of the image if the object is 10 cm away from the mirror.​

Answer 1

Answer :

The mirror is concave mirror

The magnification of the image is +1.25

The image is formed 13.33cm behind the mirror

Given :

• An image of height 2.5cm is produced by a spherical mirror
• The height of the object is 2.0cm
• The object is 10cm away from the mirror

Task :

• To identify the mirror
• To find the magnification of the image
• To find the exact location of the image

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 ,

Height of object , h1 = 2.0cm

Height of image , h2=2.5cm

Object distance, u = -10cm

Applying formula of magnification,

$\sf \implies m = \dfrac{2.5}{2} \\\\ \sf \implies m= \dfrac{4}{5} \\\\ \sf \implies m = 1.25$

Thus magnification is +1.25

Since the magnification is more than 1 so it is clear that the mirror is concave . It is because a convex mirror never produce magnification more than 1 , but a concave mirror does.

Now using mirror formula ,

$\sf \implies \dfrac{1}{v}+\dfrac{1}{-10}=\dfrac{1}{-40}\\\\ \sf \implies \dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{40} \\\\ \sf \implies \dfrac{1}{v}=\dfrac{3}{40}\\\\ \sf \implies v = \dfrac{40}{3} \\\\ \sf \implies v = 13.333cm$

Thus the image distance is 13.33cm

The positive sign indicates that the image is formed behind the mirror.