A professor reads a greeting card received on his 50th birthday with + 2.5 D glasses keeping the card 25 cm away. Ten years letter, he reads his farewell letter with the same glasses but he has to kee

Question

A professor reads a greeting card received on his 50th birthday with + 2.5 D glasses keeping the card 25 cm away. Ten years letter, he reads his farewell letter with the same glasses but he has to keep the letter 50 cm away. What should be the power (in dioptre) of the lens now so that he can keep the letter at the same distance as ten years ago?

TestHub · Accepted Answer

Firstly find out distance of retina from his eye lens i.e., image distance in first case (we are assuming distance between eye lens and retina is not changing which is not true practically we are taking that liberty to make problem more mathematically convenient. However it doesn't effect base line of problem)

$\frac{1}{v} - \frac{1}{- 25} = \frac{1}{f_{1}}$

$\frac{1}{v} + \frac{1}{25} = \frac{1}{f_{1}}$

for second case image distance should be same

$\frac{1}{v} + \frac{1}{50} = \frac{1}{f_{2}}$

$\Rightarrow \frac{1}{25} - \frac{1}{50} = \frac{1}{f_{1}} - \frac{1}{f_{2}}$

$\Rightarrow \frac{1 {cm}^{- 1}}{50} = \frac{1}{f_{1}} - \frac{1}{f_{2}} = P_{1} - P_{2} = 2 D$

Let power of his eye lens in two cases are (Pe)1, (Pe)2 respectively

$P_{1} = (2.5 + {(P_{e})}_{1})$

$P_{2} = (2.5 + {(P_{e})}_{2})$

$\Rightarrow P_{1} - P_{2} = {(P_{e})}_{1} - {(P_{e})}_{2}$

= 2 D

So loss in power is 2 D so he should use additional power of 2 D i.e. 4.5 D

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