Question

An object is placed at a distance of 25cm infront of axoncave mirror of focal length 15cm at what distance from the mirror would a screen be placed in order to obtain a sharp image ?find nature of image

Answer 1

Answer:

the image distance u = 9.3cm

that means the nature is diminished image

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.