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A person wear normal spectacles in which the distance between glasses and eyes is approximately $2 \mathrm{~cm}$, then power required is -5 D. If he wears contact lens, then the required power is
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The correct answer is:
$-4.54 \mathrm{D}$
Contact lens is more effective, so its required power is less.
For glasses:
$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{-x}-\frac{1}{\infty}=\frac{1}{f} \\ & f=-x \mathrm{~cm}=-\frac{x}{100} \mathrm{~m} \\ & \text { Power, } P=\frac{1}{f}=-\frac{100}{x}=-5 \Rightarrow x=20 \mathrm{~cm}\end{aligned}$
If he used contact lens then
$\begin{aligned} & u=-\infty, v=-22 \mathrm{~cm} \\ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{-22}=\frac{1}{f} \\ & f=-22 \mathrm{~cm}, P=-\frac{100}{22} \mathrm{D}=-4.54 \mathrm{D}\end{aligned}$
For glasses:
$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{-x}-\frac{1}{\infty}=\frac{1}{f} \\ & f=-x \mathrm{~cm}=-\frac{x}{100} \mathrm{~m} \\ & \text { Power, } P=\frac{1}{f}=-\frac{100}{x}=-5 \Rightarrow x=20 \mathrm{~cm}\end{aligned}$
If he used contact lens then
$\begin{aligned} & u=-\infty, v=-22 \mathrm{~cm} \\ & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{-22}=\frac{1}{f} \\ & f=-22 \mathrm{~cm}, P=-\frac{100}{22} \mathrm{D}=-4.54 \mathrm{D}\end{aligned}$
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