The near-point of a person suffering from hypermetropia is at 75 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).
Answers
Answered by
0
Answer:
The distance of the eye from the nearpoint is the object distance, u = -25cm
(The negative sign is placed considering the eye is in the right side)
The near point of the Hypermetropic eye is the point where image is to be formed.
so, image distance, v = -75cm
We know, for lenses
→ 1/v - 1/u = 1/f
→ 1/(-75) - 1/(-25) = 1/f
→ -1/75 + 1/25 = 1/f
→ 2/75 = 1/f
→ f = 37.5 cm = 0.375 m
Power of the lens = 1/(focal length in metres) = 1/0.375 = 2.66
Thus, the focal length is 37.5 cm & the power of the lens is 2.66. The lens is a convex lens.
#answerwithquality #BAL
Similar questions