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A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters. (i) What power lenses are required to see distant objects? (ii) What is his near point without glasses? (iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

Answers (1)

Key concepts :

$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

in terms of power $P=P_{1}+P_{2}$

(i) If for the normal relaxed eye of an average person, the power at the far point be $P_{f}$ . The required power

$P_{f}=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.02}=60\: D$

By the corrective lens, the object distance at the far point is $\infty$ .

The power required is ,

$P'_{f}=\frac{1}{f'}=\frac{1}{\infty }+\frac{1}{0.02}=50\; D$

Now for eye+lens system, we have the sum of the eye and that of the glasses $P_{g}$ .

$P'_{f}=P_{f}+P_{g}\Rightarrow 50\; D=60D+P_{g}$

Which gives, $P_{g}=-10D$

(ii) For the normal eye his power of accommodation is 4D. Let the power of the normal eye for near vision be $P_{n}$

Then, $4=P_{n}-P_{f}\; or \; P_{n}=64\; D$

Let his near point be $x_{n'}$ then

$\frac{1}{x_{n}}+\frac{1}{0.02}=64\; or\; \frac{1}{x_{n}}+50=64$

$\frac{1}{x_{n}}=14\Rightarrow x_{n}=\frac{1}{14}m=0.07 m$

(iii) With glasses $P_{n}^{'}=P{_{f}}^{'}+4=54$

$54=\frac{1}{x{_{n}}^{'}}+\frac{1}{0.02}=\frac{1}{x{_{n}}^{'}}+50$

$\frac{1}{x{_{n}}^{'}}=4\Rightarrow x{_{n}}^{'}=\frac{1}{4}m=0.25\; m$

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