Question

A shaving mirror has a radius of curvature of 30 cm. A man sees his image 2.5 times the size of his face. How far is the mirror from his face​

Answer 1

Explanation:

M = F / ( F-U)

according to sign convention

F = R/2 = 30/2 =15 cm

shaaving mirror is concave mirror

for virtual image ,M = +ve

2.5 = -15 / (-15 -U)

2.5 = 15 /(15 + U)

2.5 ( 15 + U) = 15

2.5 x 15 + 2.5 x U = 15

37.5 + 2.5 x U = 15

U =( 15 - 37.5) /2.5 = - 22.5 /2.5cm

Answer 2

Concept:-

Here, We have Radius of curvature of mirror is 30 cm and the man sees his face image is 2.5 times greater the size of his face. We have to find the distance between his face and the mirror. To find we use some certain formula:-

Given:-

• Radius of curvature of the mirror is of 30 cm.
• The size of image is 2.5 times greater than the face of man.

To Find:-

• How far is the mirror from his face ?

Solution:-

Here, We know that focal length is half of the radius of curvature.

$\sf{So,\ f = \dfrac{r}{2} \ \Longrightarrow f = \dfrac{30}{2}\ \ \Longrightarrow f = 15 cm}$

$\sf{We\ know\ that,\ m = \dfrac{-v}{u}}$

Now, It is given that

$\sf{\Longrightarrow 2.5 = \dfrac{-v}{u}}$

$\sf{\Longrightarrow 2.5u = -v}$

$\sf{\Longrightarrow v = -2.5u}$                ------------------ (i)

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

Putting all the values we get,

$\sf{\Longrightarrow \dfrac{1}{u} + \dfrac{1}{-2.5u} = \dfrac{1}{-15}}$                    [from equation (i) ]

$\sf{\Longrightarrow \dfrac{2.5 - 1}{2.5u} = \dfrac{1}{-15}}$

$\sf{\Longrightarrow \dfrac{1.5}{2.5u} = \dfrac{1}{-15}}$

By cross multiplication

$\sf{\Longrightarrow 2.5u = -15 \times 1.5}$

$\sf{\Longrightarrow 2.5u = -22.5}$

$\sf{\Longrightarrow u = \dfrac{-22.5}{2.5}}$

$\sf{\Longrightarrow u = 9\ cm}$

Hence, Mirror is 9 cm apart from the man's face.