Question

an object of height 3cm is in front of a concave mirror whose radius of curvature is 40cm, find:

1. position of image
2.magnification
3.height of image

if the object is at 60cm​

Answer 1

Given :

In concave mirror,

Height of object : 3 cm

Radius of curvature : 40 cm

Object distance : 60 cm

To find :

The position of image, magnification, height of image.

Solution :

First we have to find the focal length of the mirror,

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,

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

$\dashrightarrow \sf f = \dfrac{40}{2}$

$\dashrightarrow \sf f = 20 \: cm$

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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 60} = \dfrac{1}{ - 20}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{60} = \dfrac{1}{ - 20}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 20} + \dfrac{1}{60}$

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

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 2}{60}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 1}{30}$

$\dashrightarrow\sf v = - 30 \: cm$

Thus, the position of the image is -30 cm.

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 and it is equal to the ratio of image height and object height. that is,

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

where,

• h' denotes height of image
• h denotes object height
• v denotes image distance
• u denotes object distance

By substituting all the given values in the formula,

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

$\dashrightarrow\sf m = - \dfrac{ - 30}{ - 60}$

$\dashrightarrow\sf m = - \dfrac{1}{2}$

$\dashrightarrow\sf m = - 0.5$

Thus, the magnification of the image is 0.5.

we know that,

$\dashrightarrow\sf m = \dfrac{h'}{h}$

$\dashrightarrow\sf - 0.5 = \dfrac{h'}{3}$

$\dashrightarrow\sf h' = - 0.5 \times 3$

$\dashrightarrow\sf h' = - 1.5 \: cm$

Thus, the height of the image is 1.5 cm.