Question

A object 40cm in size is placed at 25cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image. Find the nature and size of the image.

Answer 1

Object size (O) = 40 cm

Focal length = - 15 cm

Object distance (u) = - 25 cm

Let the image distance be v.

Using the mirror formula:

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

$v = \frac{u \times f}{u-f}$

$v = \frac{15 \times 25}{-25 + 15}$

v = - 37.5 cm

The image is formed in front of the mirror and the distance is 37.5 cm.

$\frac{h_i}{h_o} = -\frac{v}{u}$

$\frac{h_i}{40}= -\frac{-37.5}{-25}$

$h_i = - 60 cm$ (image height)

The nature of the image is real, inverted and magnified.

Answer 2

Answer:

$\displaystyle \red{{ \text{h}_i=6 \ cm}$

Explanation:

Given :

Object distance u = 25 cm

Focal length f = 15 cm

We have mirror formula :

1 / f = 1 / v + 1 / u

- 1 / 15 = 1 / v - 1 / 25

1 / v = 1 / 25 - 1 / 15

5 / v = 1 / 5 - 1 / 3

5 / v = ( 3 - 5 ) / 15

5 / v = - 2 / 15

v = - 75 / 2 = - 37.5 cm

We know :

h_i / h_o = - v / u

Where i and o represent image and object

h_i = - ( - 37.5)  × 4 / 25

h_i = 6 cm

Nature of the image are as :

Real , inverted and magnified.