Question

a mirror forms an image of magnification 1/3 . if the object is placed at a distance of 10 cm from the mirror then calculate the position of the image and focal length of the mirror. ​

Answer 1

Given :

Magnification of image is 1/3

Object distance = 10cm.

To find :

The position of the image and focal length of the mirror.

Solution :

The Magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign .i.e.,

$\boxed{ \bf Magnification = - \dfrac{v}{u} }$

here,

• v denotes image distance
• u denotes object distance

Now, substituting all the given values in the formula,

$\leadsto{ \sf m = - \dfrac{v}{u} }$

$\leadsto{ \sf \dfrac{1}{3} = - \dfrac{v}{(10)} }$

$\leadsto{ \sf \dfrac{10}{3} = - v }$

$\leadsto{ \underline{ \boxed {\sf v = - 3.3 \: cm}}}$

thus, the position of the image is 3.3 cm

Now, using mirror formula,

A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

$\boxed{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }$

here,

• v denotes Image distance
• u denotes object distance
• f denotes focal length

now, substituting all the given values,

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

$\leadsto{ \sf \dfrac{1}{-10/3} + \dfrac{1}{10} = \dfrac{1}{f} }$

$\leadsto{ \sf - \dfrac{3}{10} + \dfrac{1}{10} = \dfrac{1}{f} }$

$\leadsto{ \sf \dfrac{ - 3 + 1}{10} = \dfrac{1}{f} }$

$\leadsto{ \sf \dfrac{ - 2}{10} = \dfrac{1}{f} }$

$\leadsto \underline{ \boxed{ \sf f = - 0.2 \: cm }}$

thus, the focal length of the mirror is 0.2 cm.