Question

an object of height 3cm is placed at a distance of 15cm infront of the concave mirror of radius of curvature 10cm caluculate the height of image and Image distance​

Answer 1

Given :

height of the object, h = 3cm

object distance, u = -15 cm( to the left of mirror)

Radius of curvature,R = 10cm

To find :

(a) image distance

(b) height of the image

Solution :

we know that the focal length of the spherical mirror is equal to half of its radius of curvature

f = R/2

f = 10/2 = 5cm.

Remember :- in concave mirror focal length is negative

now, using Mirror formula,

A formula which gives the relationship between image distance, object distance and focal length of a spherical mirror is known as Mirror formula.

The mirror formula can be written as,

${ \leadsto{ \bf{ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} + \dfrac{1}{ - 15} = - \dfrac{1}{5} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} - \dfrac{1}{15} = - \dfrac{1}{5} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} = - \dfrac{1}{5} + \dfrac{1}{15} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} = \dfrac{ - 3 + 1}{15} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v} = -\dfrac{2}{15} }}}$

${ \leadsto{ \bf{v = - \dfrac{15}{2} }}}$

${ \leadsto{ \bf{v = - 7.5cm }}}$

thus, image distance = -7.5cm

(b) we know that,

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

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

${ \leadsto{ \bf{ \dfrac{ - ( - 7.5)}{ - 15} = \dfrac{h'}{3} }}}$

${ \leadsto{ \bf{h' = - \dfrac{ 7.5 \times 3}{15} }}}$

${ \leadsto{ \bf{h' = - 1.5cm }}}$

height of the image is -1.5cm.

Nature of the image :

• the image is real and inverted.
• the image is diminished.