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प्रश्न
A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) most strained?
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उत्तर
(a) When the eye lens is fully relaxed, we have:
u = ∞
Distance of the retina from the eye lens, v = 2 cm = 0.02 m
The lens formula is given by
`1/v -1/u =1/f`
Putting the respective values, we get:
`1/f = 1/0.02 - 1/∞ = 1/0.02`
∴ Power of the lens =`1/f` = 50 diopters
So, in a fully relaxed condition, the power of the eye lens is 50 D.
(b) When the eye lens is most strained:
u = – 25 cm = – 0.25 m
The lens formula is given by
`1/v -1/u =1/f`
Putting the values, we get:
`1/f = 1/0.02 -1/((-0.25))`
= 50+4 =54
∴ Power of the lens = `1/f` = 54 diopters
So, in the most strained condition, the power of the eye lens is 54 D.
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The focal length of a normal eye-lens is about
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(d) If the near point is 1 m away from the eye, divergent lens should be used.
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