For a person near point of vision is 100 ~cm . Then the power of lens he must wear so as to have normal vision, should be

Correct answer: C. +3 D The power of lens P=(1)/(f)=(1)/(v)-(1)/(u)= [(1)/(25)- 1 N.P. ] where N.P. = Near point of vision aligned & P=[(1)/(25)-(1)/(100)] & P=(4-1)/(100)= 3 100 ~cm = 3 1 ~m =3 D aligned

AIIMS 2010

Exam: AIIMS
Year: 2010
Subject: Physics
Chapter: Ray Optics

AIIMS 2010 · Physics · Ray Optics

For a person near point of vision is 100 ~cm . Then the power of...

For a person near point of vision is

100 \mathrm{~cm}

. Then the power of lens he must wear so as to have normal vision, should be

Options

A $+1 \mathrm{D}$
B $-1 \mathrm{D}$
C $+3 \mathrm{D}$
D $-3 \mathrm{D}$

Correct answer

C. $+3 \mathrm{D}$

Step-by-step solution

The power of lens

P=\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=

\left[\frac{1}{25}-\frac{1}{\text{ N.P. }}\right]

where N.P. = Near point of vision

\begin{aligned} & P=\left[\frac{1}{25}-\frac{1}{100}\right] \\ & P=\frac{4-1}{100}=\frac{3}{100 \mathrm{~cm}}=\frac{3}{1 \mathrm{~m}}=3 \mathrm{D} \end{aligned}

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