Question

the near point and far point of a person are 40 cm and 250 cm respectively. determine the power of the lens he should use while reading a book kept at distance 25 cm from the eye

Answer 1

We know that , the formula is ,

1/f = 1/v - 1/u,

Normal person's Near point is 25cm,

Hypermetropic person usually can't see near objects, So what they have to do is, Use convex lens and make the object's image fall at 25 cm and let the lens of eye to make the image as object,

So what we have to do here is, Make the near point of Hypermetropic person to 25cm,

So here object distance = Near point of Hypermetropic person,

Image distance = -25cm(- indicates direction),

=> Object distance = u = -40cm (Direction),

Focal length of Convex lens is always positive !,

Now Inputting all the values to find focal length in cm,

=> 1/f = -(1/25) -(-1/40),

=> 1/f = 1/40 - 1/25,

=> 1/f = (5-8)/200

=> 1/f = -3/200 cm,

=> f = -200/3 cm, Converting it into m,

=> f = -200/3/100 = -2/3 m,

So therefore focal length = -2/3 m,

Now calculating power,

Power = 1/Focal length in m ,

Units = D , Dioptre,

=> Power = 1/-2/3 = -3/2D

Answer 2

Answer:

The power of the lens he should use while reading a book kept at a distance $25$ cm from the eye is $-1.5D$.

Explanation:

The lens should be such that it shifts the image of an object at $25$ cm from the eye to the near point of the defected person.

Hence the lens would follow the following lens equation,

$\bold{\frac{1}{f} =\frac{1}{v} +\frac{1}{u}}$

Where,

$v=$ the nearest point of the eye

$u=$ the far point of the eye

$f=$ focal length

Now substitute all the given values:

$\frac{1}{f}=\frac{1}{-0.4} -\frac{1}{-0.25} \\\frac{1}{f}=-\frac{1}{0.4} +\frac{1}{0.25} \\\frac{1}{f}=-1.5D$

Hence, the power of the lens is $-1.5D$