Question

The nearer point of hypermetropic eye is 40 cm. The lens to be used for its correction should have the power?
A. +1.5 D.
B. -1.5 D.
C. +2.5 D.
D. +0.5 D.

Answer 1

Hint: To solve this question, we have to remember that the near point of a hypermetropic eye is more that 25 cm away. Such a person has to hold the reading material much beyond 25 cm from eye for comfortable reading.

Complete answer:
Given that,
Nearer point of hypermetropic eye = 40 cm.
As we know that the hypermetropia defect of the eye is corrected by using a convex lens. So, the person requires convex lens spectacles.
First, we will calculate the focal length for that lens.
We know that the hypermetropic eye can see the nearby object kept at 25 cm clearly if the image of this object is formed at its own near point which is 40 cm here.
So, we have
Object distance, u = -25 cm [according to sign convention of convex lens]
Image distance, v = -40 cm [according to sign convention of convex lens]
We will use the lens formula to find the focal length of this lens,
The lens formula is given by,
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ …… (i)
Putting the values in equation (i), we will get
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{ - 40}} - \dfrac{1}{{ - 25}}$
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{ - 40}} + \dfrac{1}{{25}}$
$ \Rightarrow \dfrac{1}{f} = \dfrac{{ - 5 + 8}}{{200}}$
$ \Rightarrow \dfrac{1}{f} = \dfrac{3}{{200}} = \dfrac{1}{{66.66}}cm$
Hence, we get focal length, f = 66.66 cm = $\dfrac{{66.66}}{{100}}m$= 0.6666m.
Now, we will calculate the power of the lens.
The power of lens can be obtained by,
Power, P = $\dfrac{1}{{{\text{focal length of the lens (in metres)}}}}$
So,
P = $\dfrac{1}{{0.6666}}$
P = +1.5 D.
Hence, the lens to be used for its correction should have the power of +1.5 D

So, the correct answer is “Option A”.

Note: The convex lens used for correcting hypermetropia (or long-sightedness) should be of such a focal length (or power) that it forms a virtual image of the object (placed at a normal near point of 25 cm), at the near point of the hypermetropic eye. The defect of hypermetropia is caused due to low converging power of the eye lens or due to small eyeballs.