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**Solutions: Medical Option - the EYE**

**Q9. **

A defective eye has an unaided near point at 0.65 m and an unaided far point at infinity.

Calculate

(i) the power of the correcting lens needed to allow the eye to see clearly an object 0.25 m from the eye,

v = - 0.65 m

u = 0.25 m

P = 1/f = 1/v + 1/u

= -1/0.65 + 1/0.25 = 2.46

= + 2.5 D

(ii) the furthest distance from the eye that an object can be seen clearly when the correcting lens is used.

The furthest distance the image can be formed is at infinity for the eye - therefore 1/v will be zero..

P = 1/u + 1/v

2.46 = 1/u + 0

u = 0.41 m

**(Total 3 marks) **