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A young boy can adjust the power of his eye-lens between 50 D and 60 D. His far points is infinity. (a) What is the distance of his retina from the eye-lens? (b) What is his near point? |
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Answer» Solution :a) When the EYE is fully relaxed, its focal length is largest and the power of the eye-lens in minimum. This power is 50 D according to the GIVEN data. The focal length is `1/50`m = 2cm. As the far point is at INFINITY, the parallel rays COMING from infinity are focused on the retina in the fully relaxed condition. Hence, the distance of the retina from the lens equals the focal length which is 2 cm. When the eye is focused at the near point, the power is maximum which is 60 D. The focal length in this case is `f=1/60m=5/3cm`. The image is formed on the retina 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 = -10 cm. The near points is at 10 cm. |
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