An object 4cm in size, is placed at 25cm infront of a concave mirror
of focal length 15cm. At what distance from the mirror whould a screen be
placed in order to obtain a sharp image?Find the nature and the size of the
image.

 

Asked by amarapurevathi | 5th Oct, 2018, 12:22: PM

Expert 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. 

Answered by Shiwani Sawant | 5th Oct, 2018, 02:25: PM