A person wear normal spectacles in which the distance between glasses and eyes is approximately 2 ~cm , then power required is -5 D. If he wears contact lens, then the required power is

Correct answer: B. -4.54 D Contact lens is more effective, so its required power is less. For glasses: aligned & (1)/(v)-(1)/(u)=(1)/(f) (1)/(-x)-(1)/(∞)=(1)/(f) & f=-x cm=-(x)/(100) m & Power, P=(1)/(f)=-(100)/(x)=-5 x=20 cm aligned If he used contact lens then aligned & u=-∞, v=-22 cm & (1)/(v)-(1)/(u)=(1)/(f) (1)/(-22)=(1)/(f) & f=-22 cm, P=-(100)/(22) D=-4.54 D aligned

AIIMS 201925 May

Exam: AIIMS
Year: 2019
Date: 25 May
Subject: Physics
Chapter: Ray Optics

AIIMS 2019 · 25 May · Physics · Ray Optics

A person wear normal spectacles in which the distance between...

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

Options

A $-5.2 \mathrm{D}$
B $-4.54 \mathrm{D}$
C $+5.2 \mathrm{D}$
D $+4.7 \mathrm{D}$

Correct answer

B. $-4.54 \mathrm{D}$

Step-by-step solution

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}

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