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A young boy can adjust the power of his ...

A young boy can adjust the power of his eye lens between 50 D and 60 D. His far point is infinity. A. What is the distance of his retina from the eye-lens? b. What is the near point?

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a. When the eye is fuly relaxed, its focal length is largest and the power of the eye lens is minimum. This power is 50 D according to the given data. The focal lengts is `1/50m=2cm.` As the far point is at infinity, the parallel rays coming from infinity are focussed on the retiN/A in the fuly relaxed condition. HEnce, the distance of the retinN/A from the lens equals the focal length which is 2 cm.
b. when the eye is focussed at the near point, the power is maximum which is 60 D. the focal lenth in this case is `f=1/60m=5/3cm. The image is formed on the retiN/A and thus v=2 cm. We have,
`1/v-1/u=1/f`
or `1/u=1/v-1/f=1/(2cm)-3/(5cm)`
or `u=-10cm`
The near point is at 10 cm
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