Question

A concave mirror produces 5 times enlarged image, of an object placed at 15 cm from the mirror. Find the focal length of mirror and position of image if
1 image formed is real,
2 image formed is virtual.​

Answer 1

Given :

In concave mirror,

Magnification of the image = 5

Object distance = - 15 cm

To find :

The focal length of mirror and position of image if

• Image formed is real,
• Image formed is virtual.

Solution :

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

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

By substituting all the given values in the formula

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

$\dashrightarrow \sf 5 = - \dfrac{v}{ - 15}$

$\dashrightarrow \sf v = 5 \times 15$

$\dashrightarrow \boxed{\sf v = 75}$

If the image is virtual then it denoted by minus sign and positive sign if it is real.

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.,

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

where,

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

By substituting all the given values in the formula,

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

$\dashrightarrow\sf \dfrac{1}{75} + \dfrac{1}{ - 15} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{75} - \dfrac{1}{ 15} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{5 - 1}{75} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{f} = \dfrac{4}{75}$

$\dashrightarrow\sf f = \dfrac{75}{4}$

$\dashrightarrow \boxed{\sf f = 18.75 \: cm}$