InterviewSolution
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InterviewSolution
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The near and far points of a person are at `40 cm` and `250cm` respectively. Find the power of the lens he/she should use while reading at `25cm`. With this lens on the eye, what maximum distance is clearly visible? |
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Answer» If an object is placd at 25 cm from the correcting lensl it should produce the virtual image at 40 cm. Thus, `u=-25cm, v=-40cm` `1/f=1/v-1/u` `=1/(40cm)+1/(25cm)` or `f=200/3cm=+2/3m` `P=1/f+1.5D` The uN/Aided eye can see a maximum distance of 250 cm. Suppose the maximum distance for clear vision is d when the lens is used. Then the object at a distance d is imaged by the lens ast 250 cm. We have `1/v-1/u=1/f` `or - 1/(250cm)-1/d=3/(200cm)` `or d=-53cm` Thus the person will be able to see up to a maximum distance of 53 cm. |
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