Question

A convex mirror of focal length 25cm, is used as a back view mirror of a car. If a another car is at a distance of 200 cm from the mirror, find
1 position of image,
2 magnification of image.​

Answer 1

Given :

In convex mirror,

Focal length = 25 cm

Object distance = - 200 cm

To find :

The position of image and magnification of image.

Solution :

using mirror formula that is,

» 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 200} = \dfrac{1}{25}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{ 200} = \dfrac{1}{25}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{25} + \dfrac{1}{ 200}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{8 + 1}{ 200}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{9}{ 200}$

$\dashrightarrow\sf v = \dfrac{200}{9}$

$\dashrightarrow\sf v = 22.2 \: cm$

Thus, the position of the image is 22.2 cm.

» The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign that is,

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

where,

• m is magnification
• v is image distance
• u is object distance

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

$\dashrightarrow \sf m = -\dfrac{ \dfrac{200}{9} }{ - 25}$

$\dashrightarrow \sf m =\dfrac{200}{ 9 \times 25}$

$\dashrightarrow \sf m =\dfrac{8}{ 9 }$

$\dashrightarrow \sf m =0.8$

Thus, the magnification of the image is 0.8