Question

An object 4.0 cm in size is placed at 25.0cm infront of a concave mirror of focal length 15.0cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.

Answer 1

Explanation:

Given:

Object-size,h=+4.0cm;

Object-distance,u=-25.0cm;

Focal length,f=-15.0cm;

We want to find:

Image-distance,v and Image-size,h'

FORMULA:

1/v+1/u=1/f

ATQ,1/v=1/f-1/u

1/v = (1/-15.0) - (1/-25.0)

1/v = (-1/15.0)+(1/25.0)

1/v = (-25+15)/375

1/v = -10/375

1/v = -2/75

v = -37.5

The screen should be placed at 37.5cm in front of the mirror.The image is real.

Also,magnification,m=h'/h= -v/u

ATQ,

h'= -vh/u

= -(-37.5cm) (+4.0cm)/(-25.0cm)

Height of the image,h'= -6.0cm

The image is inverted and enlarged