19. The near point of a hypermetropic eye is at 75 cm from the eye. What is the power of the lens required to enable him to read clearly a book held at 25 cm from the eye?
Answers
Answer:The near point of the eye is 75 cm which means the person is not able to see any objects before 75 cm. So, we can say that the person is suffering from hypermetropia. So, the lens to be used is the convex lens.
What this will do is bring the near point from 75 cm to 25 cm as stated in the question. So, any object placed at 25 cm should be viewed clearly by the person.
Object distance is u = -25 cm
The lens will form a virtual image at 75 cm (the distance at which the person can see). So, v = -75 cm (negative because virtual image)
begin mathsize 12px style 1 over straight f equals 1 over straight v minus 1 over straight u
1 over straight f equals fraction numerator 1 over denominator negative 75 end fraction minus fraction numerator 1 over denominator negative 25 end fraction equals 1 over 25 minus 1 over 75 equals 2 over 75
straight f equals 75 over 2 equals 37.5 space cm end style
Therefore, power of lens is
begin mathsize 12px style straight P equals fraction numerator 1 over denominator straight f space left parenthesis in space straight m right parenthesis end fraction equals fraction numerator 1 over denominator 0.375 straight m end fraction equals plus 2.67 space straight D end style