Question

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describethe nature and size of the image. If the candle is moved closer to themirror, how would the screen have to be moved?

Answer 1

Given: The size of the candle (object) is 2.5 cm, the distance of the candle (object) from the mirror is 27 cm and the radius of curvature is 36 cm.

The formula of the mirror is given as,

1 u + 1 v = 1 f 1 u + 1 v = 1 ( R 2 )       [ ∵f= R 2 ] 1 u + 1 v = 2 R

Where, the distance of the object is u, the distance of the image is v, the radius of curvature is R and the focal length is f.

By substituting the values in the above formula, we get

1 −27 + 1 v = 2 −36 1 v = 1 −18 − 1 −27 1 v =− 1 54 v=−54 cm

The negative sign indicates that the image is formed in front of the mirror.

Thus, the image is formed at 54 cm at the front of the mirror.

The magnification is given as,

m= −v u h i h o = −v u

Where, the object height is h o and the image height is h i .

By substituting the values in the above expression, we get

h i 2.5 =−( −54 −27 ) h i =−5 cm

Thus, the image formed is inverted and real. As, the candle is moved closer to mirror the screen would have to be moved farther.

When the candle closer than 18 cm from the mirror, the image would be virtual and therefore cannot be collected on the screen.