Question

A concave mirror produces four times magnified real image of a real object . If the distance between object and image is 60 cm , then find the value of radius of curvature .

Answer 1

Answer:

19.3 cm

Explanation:

Given :

Magnification = 4

i.e. 4 = v / - u

= > v = - 4 u

Difference is 60 cm

i.e. u - v = 60 cm

Putting v = - 4 u

u + 4 u = 60

u = 12 cm

So , v = - 4 × 12 = > - 48 cm

We have to find radius of curvature .

First finding focal length by using mirror formula :

1 / f = 1 / u + 1 / v

1 / f = 1 / - 12 + 1 / - 48  [ Used sign convention ]

1 / f = - 5 / 48

f = - 48 / 5 cm

f = 9.6 cm  [ Negative sign shows direction only ]

Now :

R = 2 f

R = 19.3 cm

Hence radius of curvature is 19.3 cm

Answer 2

Solution:

Given:

=> Magnification = 4

=> Distance between object and image is 60 cm

To Find:

=> Radius of curvature.

Formula used:

$\sf{\implies \dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}\;\;\;\;[Mirror\;Formula]}$

$\sf{\implies R = 2f}$

Now, magnification = 4

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

=> v = -4u     ...........(1)

And it is given that distance between object and image is 60 cm. So,

=> u  - v = 60     .........(2)

Now, put the value of v from Equation (1) to Equation (2), we get

=> u - (-4u) = 60

=> u + 4u = 60

=> 5u = 60

=> u = 60/5

=> u = 12 cm

Now, put the value of u in Equation (1), we get

=> v = -4u

=> v = -4 × 12

=> v = -48 cm

Now, we will find focal length by mirror formula,

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

$\sf{\implies \dfrac{1}{f}=\dfrac{1}{-12} +\dfrac{1}{-48}}$

$\sf{\implies \dfrac{1}{f} = \dfrac{4+1}{-48}}$

$\sf{\implies \dfrac{1}{f}=-\dfrac{5}{48}}$

$\sf{\implies f = 9.6\;cm}$

Now, we will find radius of curvature,

=> R = 2f

=> R = 2 × 9.6

=> R = 19.3 cm

∴ Hence, radius of curvature is 19.3 cm