Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answers
Answered by
4
here is the answer of the correction of hypermetropia.please make me brainliest.rest parts don't know .
Attachments:
snigdhapoddar01:
whose handwriting is this?
Answered by
5
Answer:
Near point of defective eye is 1 m and that of normal eye is 25 cm.
Here, u = -25 cm, v = -1m = 100 cm.
Using lens formula
1/f = 1/v – 1/u
1/f = 1/-100 + 1/25 = 3/100
f = 100/3 cm = 1/3m.
P = 1/f(m) = 1/0.33 = +3.0 D.
Attachments:
Similar questions