Physics, asked by mnlprasath, 1 year ago

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

Answered by siril
154

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!!

Attachments:

mnlprasath: but how image distance is negative
siril: wait i'll edit and attach a picture
mnlprasath: thank you
siril: refresh the page.....the negative X-direction is represented with X'..
mnlprasath: ok according to this image formed in positive x side so the image distance must be positive is it so ? pls answer
siril: yes and the image distance or object distance is to be from the eye
siril: since it is to the left side, it is negative here
Answered by VishalSharma01
148

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.

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