Question

An object 4cm 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

Answer:

f= -15

u =-25

ho= 4

1/f=1/v+1/u

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

1/v=25-15/-375

v=-37.5

m=-v/u=hi/ho

m=-(-37.5)/-25=hi/4

hi=-6cm

Explanation:

mirror should a screen be placed at 37.5 cm in order to obtain a sharp image

nature is real, magnified and inverted

size of image is 6 cm

Answer 2

Answer:

focal length = f = - 15 cm

object distance = u = -25 cm

image distance = v = ?

hi = ?

ho = 4 cm

Mirror formula,

1/v + 1/u = 1/f

1 over f space equals space 1 over u plus 1 over v

rightwards double arrow fraction numerator 1 over denominator negative 15 end fraction equals fraction numerator 1 over denominator negative 25 end fraction plus 1 over v

rightwards double arrow space fraction numerator 1 over denominator negative 15 end fraction plus 1 over 25 space equals space 1 over v

rightwards double arrow space 1 over v equals fraction numerator 25 minus 15 over denominator negative 375 end fraction

rightwards double arrow space v space equals negative space 37.5 space c m space

m equals fraction numerator negative v over denominator u end fraction equals h subscript i over h subscript o

rightwards double arrow space m space equals fraction numerator negative left parenthesis negative 37.5 right parenthesis over denominator negative 25 end fraction equals h subscript i over 4

h subscript i space equals space 150 over 25 equals space minus 6 space c m

Thus, to get image of object sharp it has to be placed at 37.5 cm

Size of image is 6 cm

As the height of object is more and negative, nature of image is real, inverted and magnified