Question

i) Write sign convention for reflection by spherical mirrors.
ii) An object, 5.0 cm in size, is placed at 30 cm in front of a concave mirror of
focal length 15.0 cm. 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

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

Explanation: