Question

An object of height 1.2m is placed before a concave mirror of focal length 20cm so

that a real image is formed at a distance of 60cm from it. Find the position of an

object. What will be the height of the image formed?​

Answer 1

$\sf{\underline{Given:}}$

Height of the object (h) = 1.2 m = 120 cm

Focal length (concave mirror) (f) = - 20 cm

Image distance (v) = - 60 cm

$\sf{\underline{Formula\:to\:be\:used:}}$

$\boxed{\sf{\frac{1}{f} = \frac{1}{u} + \frac{1}{v} }}$

$\sf{\underline{Note:}}$ Above equation is applicable for both the mirrors (convex and concave).

$\sf{\frac{1}{u} = \frac{1}{f} - \frac{1}{v} }$

$\sf{\frac{1}{u} = - \frac{1}{20} + \frac{1}{60} }$

$\sf{ \frac{1}{u} = - \frac{1}{30} }$

$\boxed{\sf{u = - 30 \: cm}}$

For height of the image (concave mirror),

$\sf{\underline{Use\:this\:formula:}}$

$\boxed{\sf{m = \frac{ - v}{u} = \frac{h^{'}}{h}}}$

$\sf{\underline{Here:}}$

$\sf{h = 1.2 \: m = 120 \: cm}$

$\sf{v = - 60 \: cm}$

$\sf{u = - 30 \: cm}$

$\sf{\underline{Therefore:}}$

$\sf{ \frac{ - ( - 60)}{ - 30} = \frac{h^{'} }{120}}$

$\sf{ \frac{120 \times 60}{ - 30} = h^{'} }$

$\sf{ \frac{20 \times 12}{ - 1} = h^{'} }$

$\sf{ - 240 \: cm = h^{'} }$

$\sf{ - 2.4 \: m = h^{'}}$

$\boxed{\sf{h^{'} = - 2.4\:m} }$

The negative sign indicates inverted image.

Answer 2

Answer:

u = 30 cm

$\display \text{h _i = 2.4 cm}$

Explanation:

We have given :

Focal length = 20 cm

Image distance =  60 cm

We have to find object distance .

We know :

1 / f = 1 / v + 1 / u

Since image formed is real :

f = - 20 cm and v = - 60 cm

1 / u = 1 - 3 / 60

u = - 30 cm

Hence object distance is 30 cm .

Also given object height = 1.2 cm

We know :

h_i / h_o = - v / u

h_i / 1.2 = - 60 / 20

h_i = - 2.4 cm

Hence image height is 2.4 cm .