Calculate
the power of eye lens when ciliary muscles are fully relaxed??
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A normal eye can see objects from 25cm (near point) till infinity (far point).
So, we take:
Object distance, u = -∞ (- infinity)
Image distance, v = 25 cm
By using lens formula, we get,
1/v - 1/u = 1/f
1/25 - 1/-∞ = 1/f
1/f = 1/25 + 0 [Since, 1/-∞ = 0]
1/f = 1/25
f = 25 cm
For determining the power of the lens, focal length should be in metres.
Therefore, 25/100 = 0.25 m
Power, P = 1/f
P = 1 / 0.25
P = 4 D
Therefore, the power of a normal eye is 4 Dioptres.
Hope this helps u...........
So, we take:
Object distance, u = -∞ (- infinity)
Image distance, v = 25 cm
By using lens formula, we get,
1/v - 1/u = 1/f
1/25 - 1/-∞ = 1/f
1/f = 1/25 + 0 [Since, 1/-∞ = 0]
1/f = 1/25
f = 25 cm
For determining the power of the lens, focal length should be in metres.
Therefore, 25/100 = 0.25 m
Power, P = 1/f
P = 1 / 0.25
P = 4 D
Therefore, the power of a normal eye is 4 Dioptres.
Hope this helps u...........
rima21:
But when they are fully relaxed then their is maximum stretch of eye lens
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