Question

A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror,
find the position, nature and size of the image.​

Answer 1

Given :

In convex mirror,

Radius of curvature : 3 m

Object distance : - 5 cm

To find :

The position, magnification and nature of the image.

Solution :

First we have to find focal length we know that,

» For a spherical mirror having small aperture, the principle focus lies exactly mid way between the pole and centre of curvature. So, the focal length of a spherical mirror is equal to the half of its radius of curvature.

if f is the focal length of a mirror and R is its radius of curvature, then f = R/2

by substituting the given values in the formula,

$\dashrightarrow \sf f = \dfrac{R}{2}$

$\dashrightarrow \sf f = \dfrac{3}{2}$

$\dashrightarrow \sf f = 1.5 \: m$

Now, 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 5} = \dfrac{1}{1.5}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{5} = \dfrac{1}{1.5}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{1.5} + \dfrac{1}{5}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{10 + 2}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{12}{15}$

$\dashrightarrow\sf v = \dfrac{15}{12}$

$\dashrightarrow\sf v = 1.25 \: m$

Thus, the position of the image is 1.25 m.

we know that,

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

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

By, substituting all the given values in the formula,

$\dashrightarrow\sf m = \dfrac{ - 1.25}{- 5}$

$\dashrightarrow\sf m = 0.25$

Thus, the magnification of the image is 0.25

Nature of the image :

• The image is erect and virtual.
• The image is diminished.

Answer 2

Answer:-

• Radius of the Curvature=3m
• Object distance=-5m=u

$\boxed{\sf Focal\:Length=\dfrac{R}{2}}$

$\\ \tt\hookrightarrow f=\dfrac{3}{2}$

$\\ \tt\hookrightarrow f=1.5m$

According to mirror formula

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

$\\ \tt\hookrightarrow \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

$\\ \tt\hookrightarrow \dfrac{1}{1.5}-(\dfrac{1}{-5})$

$\\ \tt\hookrightarrow \dfrac{1}{1.5}+\dfrac{1}{5}$

$\\ \tt\hookrightarrow \dfrac{10+2}{15}$

$\\ \tt\hookrightarrow \dfrac{12}{15}$

$\\ \tt\hookrightarrow v=\dfrac{15}{12}$

$\\ \tt\hookrightarrow v=1.25m$

Now

$\boxed{\sf magnification=-\dfrac{v}{u}}$

$\\ \tt\hookrightarrow \dfrac{-1.25}{-5}$

$\\ \tt\hookrightarrow 0.25$

Hence

• Image is situated 1.25m from the mirror
• Image is Erect and Virtual.
• Image is diminished