Q. 7 :- Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. 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:-
As we know that hypermetropia is corrected by using a convex lens.
Here, The near point of a hypermetropic eye is 1 m and the near point of the normal eye is 25 cm.
Therefor, v= − 1 metre = − 100 cm and u= −25 cm
As per lens formula :
| 1 | 1 | 1 | |||
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− |
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| v | u | f | |||
| 1 | 1 | 1 | |||
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− |
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| − 100 | − 25 | f | |||
| − 1 + 4 | 1 | ||||
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− |
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|||
| 100 | f | ||||
| 100 | |||||
| f | = |
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cm | ||
| 3 | |||||
| 1 | |||||
| f | = |
|
m | ||
| 3 | |||||
| 1 | |||||
| P | = |
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|||
| f | |||||
| 1 × 3 | |||||
| P | = |
|
Dioptre | ||
| 1 | |||||
| P | = | + 3.0 | D | ||
