Question

An object 5.0 cm in length is placed at
a distance of 20 cm in front of a convex
mirror of radius of curvature 30 cm. Find
the position of the image, its nature and
size.
​

Answer 1

Let's find position, nature and size of the image

Given that,

Object height = 5 cm

Object distance from mirror = 20 cm

Radius of curvature = 30 cm

We know that focal length equal to half of radius of curvature. Hence focal length = 15 cm

1) = Position of image

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ \frac{-1}{15} = \frac{1}{v} + \frac{1}{ - 20} \\ \\ \frac{1}{v} = \frac{-1}{15} - \frac{1}{ - 20} \\ \\ \frac{1}{v} = \frac{-4 + 3}{60} \\ \\ v = -60cm$

(2) = nature of image

Image distance ( v ) value negative indicates that image is virtual

(3) = Its size

$m \: = \frac{hi}{ho} = \frac{ - v}{u} \\ \\ \frac{hi}{5} = \frac{ - 60}{ 20} \\ \\ hi \: = -15 \: cm$

Hence the size of image is -15 cm negative sign indicates down to the principal axis

Thank you

Answer 2

Answer:

Radius of curvature (R) = 30 cm

f = R/2 = 30/2 = 15 cm

u = --20 cm, h= 5 cm.

1/v +1/u = 1/f

1/v = 1/15+ 1/20 = 7/60

v = 60/7 = 8.6 cm.

image is virtual and erect and formed behind the mirror.

hi/h0= v/u

hi/5= 8.6/20

hi = 2.2 cm.

Size of image is 2.2 cm.