Question

A concave mirror produces a real, four times magnified image of an object placed at 10cm infront of it.Find the position of the image and the radius of curvature of the mirror.

Answer 1

Let u be the object distance and v be the image distance and the magnification of the mirror be represented by m.

We know, f is the focal length.

Given,

Its a concave mirror image should be real and inverted , magnification as in

-ve.

m = -4  

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

$v=4u$

 

u = object distance from the mirror = 10 cm = -10 cm (concave)

$v=4*(-10) =-40 cm$

Thus the image formed is 40 cm in front of the mirror.

For Position :

u = -10 cm  

v = -40 cm

We know Mirror Formula,

on putting the values,

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\\\frac{1}{f}=\frac{1}{(-40)}+\frac{1}{(-10)}\\\\\frac{1}{f}=\frac{1}{(-8)}(on solving )\\\\f =-8cm.$

The radius of Curvature , R = 2 F

                                             = 2 * (-8)

                                             = -16 cm.

As the mirror is concave -ve sign is must.

Answer 2

$\huge\boxed{SOLUTION}$

• The radius of curvature of the mirror is 16 cm and The position is left of the mirror.

• Radius = 16 cm
• Object distance = 10cm
• M²/M¹ = 4

According to given question:-

$m = \frac{ - v}{u}$

$4 = \frac{ - v}{10}$

$v = - 40cm$

USING MIRROR FORMULA

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

$= ( \frac{1}{f} = \frac{1}{10} + \frac{1}{ - 46} )$

$= ( \frac{1}{f} = \frac{1 - 4}{40} )$

$= ( \frac{1}{f} = \frac{ - 5}{46} )$

$= f = - 8cm$

• focal Length = -8cm
• Radius = -16cm

-8 cm unit is the unit and focal length.