the near point of a person suffering from hypermetropia is 75cm calculate thefocal length and power of lens required to enable him to read news paper which is kept at 25cm from the eye?
Answers
The distance of the eye from the newspaper 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
Hope it helps!!
Answer:
Explanation:
Given :-
u = - 25 cm
v = - 75 cm
To Find :-
f = ??
P = ??
Solution :-
Using, 1/v - 1/u = 1/f we get
⇒ 1/- 75 - 1/- 25 = 1/f
⇒ 1/- 75 + 1/25 = 1/f
⇒ 2/75 = 1/f
⇒ f = 75/2
⇒ f = 37.5 cm
Hence, the focal length is 37.5 cm.
Now, P = 100/37.5 cm
⇒ P = 2.6 D
Hence, the power of lens required to enable him to read news paper is 2.6 D.