Question

A hypermetropia is corrected for near vision by using a lens of power 8/3

d. Calculate how far is near point of defective eye

Answer 1

Answer:

hypermetric : A person suffering from hypermetropic can see distinct objects clearly but faces difficulty in seeing nearby objects. it means, when we places an object at 25cm(nearest point of normal eyes ) , person will see object' image at x cm, this is known as near point of defective eye.

here, object distance , u = -25cm

image distance, v = x cm

power of lens, P = +8/3 D

so, focal length , f = 1/P = 3/8 m or, 300/8 = 75/2 cm

now, using formula 1/v - 1/u = 1/f

or, 1/x - 1/-25cm = 2/75

or, 1/x = 2/75 - 1/25 = -1/75

or, x = -75 cm

hence, near point of defective eye is 75cm far.

hope it helps you my friend

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