Question

A convex mirror used a rear-view mirror in a car has a radius of curvature of 3 m. If a bus is located at a distance of 20m from this position of image. What is the nature of image ?

Answer 1

Given :

In convex mirror,

Radius of curvature = 3m

Object distance = 20 m

To find :

The position of the image and nature of the image.

Solution :

1st 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 20} = \dfrac{1}{1.5}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{ 20} = \dfrac{1}{1.5}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{1.5} + \dfrac{1}{ 20}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{40 + 3}{ 60}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{43}{ 60}$

$\dashrightarrow\sf v = \dfrac{60}{43}$

$\dashrightarrow\sf v = 1.39 \: m$

Thus, the position of the image is 1.39 m.

Hence, the nature of image is the image is erect and virtual.

Answer 2

Hola ⚘⚘

$Given :-$

• Radius of curvature = 3m
• Object distance = 20 m

$To find :-$

• Position of the image
• Nature of the image.

$Solution :-$

Finding the focal length

$\implies\:\: f = \dfrac{R}{2}$

$\implies\:\: f = \dfrac{3}{2}$

$\implies\:\:f = 1.5 \: m$

Now, Using mirror formula

$\implies\:\: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

v = Image distance

u = object distance

f = focal length

$\implies\:\: \dfrac{1}{v} + \dfrac{1}{ - 20} = \dfrac{1}{1.5}$

$\implies\:\: \dfrac{1}{v} - \dfrac{1}{ 20} = \dfrac{1}{1.5}$

$\implies\:\: \dfrac{1}{v} = \dfrac{1}{1.5} + \dfrac{1}{ 20}$

$\implies\:\: \dfrac{1}{v} = \dfrac{40 + 3}{ 60}$

$\implies\:\: \dfrac{1}{v} = \dfrac{43}{ 60}$

$\implies\:\: v = \dfrac{60}{43}$

$\implies\:\: v = 1.39 \: m$

• Position of the image is 1.39 m

• Nature of image erect and virtual