Question

a convex mirror used for rare view on the automobile has a radius of curvature 3 m by the bus is located 5 m from his mirror .find the position nature size of object​

Answer 1

Given :

In convex mirror,

Radius of curvature = 3m

Object distance = 5 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,

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

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

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

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

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{10 + 3}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{13}{15}$

$\dashrightarrow\sf v = \dfrac{15}{13}$

$\dashrightarrow\sf v = 1.15 \: m$

Thus, the position of the image is 1.15 m.

Thus, the position of the image is 1.15 m.Hence, the nature of image is the image is erect and virtual.

Answer 2

Bonjour ⚘⚘

$Given :-$

• Radius of curvature = 3m
• Distance of object = 5 m

$Tofind :-$

• Position of the image
• Nature of the image

$Solution :-$

Firstly, let's find 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

$\implies\:\:\: \dfrac{1}{v} + \dfrac{1}{ - 5} = \dfrac{1}{1.5}$

$\implies\:\:\: \dfrac{1}{v} - \dfrac{1}{5} = \dfrac{1}{1.5}$

$\implies\:\:\: \dfrac{1}{v} = \dfrac{1}{1.5} + \dfrac{1}{5}$

$\implies\:\:\: \dfrac{1}{v} = \dfrac{10 + 3}{15}$

$\implies\:\:\: \dfrac{1}{v} = \dfrac{13}{15}$

$\implies\:\:\: v = \dfrac{15}{13}$

$\implies\:\:\: v = 1.15 \: m$

• Position of the image is 1.15 m

• Nature of image is erect and virtual