Question

The near point of a short-sighted person is 10 cm and he desires to read a book 30 cm away from him. The power of the lens to be used by him is: (1) –3.33 D (2) –10 D (3) –6.66 D (4) –5 D

Answer 1

Answer:

the answer is a option C

Answer 2

Answer:

The power of the lens to be used by him is -6.66 D i.e.option(3).

Explanation:

We will use the lens formula to solve this question. Which is given as,

$\mathrm{\frac{1}{v} -\frac{1}{u} =\frac{1}{f} }$       (1)

Where,

v=image distance from the person's eye

u=object distance from the person's eye

f=focal length

From the question we have,

u = near point of short-sighted person = -10 cm

v =image distance= -30cm

By inserting the entities in equation (1) we get;

$\mathrm{\frac{1}{-30} -\frac{1}{-10} =\frac{1}{f} }$

$\mathrm{\frac{1}{-30} +\frac{1}{10} =\frac{1}{f} }$

$\mathrm{\frac{1}{15}= \frac{1}{f}}$

$\mathrm{f=15\ cm}$      (2)

The power of the lens is calculated as,

$\mathrm{P=-\frac{100}{f}}$        (3)

Inserting the value of focal length in equation (3) we get;

$\mathrm{P=-\frac{100}{15}}$

$\mathrm{P=-6.66\ D}$

The power of the lens to be used by him is -6.66 D i.e.option(3).

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