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    				| 1. | Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm. | 
| Answer» SOLUTION :The diagram is as shown below   `x=1m=100cm` `d=25cm` `f=(xd)/(x-d)=(100xx25)/(100-25)` `f=33.33cm` `P=(1)/(f(m))` `=(100)/(f(CM))=(100)/(33.33)=+3D` | |