Question

A farsighted person has his near point $50\,{\text{cm}}$ . Find the power of the lens he should use to see at $25\,{\text{cm}}$ clearly.
A. $+ 1\,{\text{D}}$
B. $+ 2\,{\text{D}}$
C. $- 2\,{\text{D}}$
D. $- 1\,{\text{D}}$

Answer 1

First of all, we will identify the object and image distance from the numerical. We will use the lens formula to calculate the focal length in meters. We will then take the inverse of focal length to find out the power.

Complete step by step answer:
In the given question, the following data are supplied:
The near point of a far-sighted person is given as $50\,{\text{cm}}$ .
The person has no difficulty in viewing objects placed at a distance, as this is the case of farsightedness.
There should be a lens for which he would be able to see an object which is placed less than $50\,{\text{cm}}$ i.e. $25\,{\text{cm}}$ .
So, we consider the image distance to be $50\,{\text{cm}}$ .
The object distance to be$25\,{\text{cm}}$ .
We need to calculate the power of the lens, for which we need to first calculate the focal length.

We can write,
$v = - 50\,{\text{cm}}$
$u = - 25\,{\text{cm}}$

We apply the lens formula,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ …… (1)
Where,
$v$ indicates the distance of the image.
$u$ indicates the distance of the object.
$f$ indicates the focal length.

Now, we substitute the required values in the equation (1), we get:
\dfrac{1}{{ - 50}} - \dfrac{1}{{ - 25}} = \dfrac{1}{f} \\\
\implies \dfrac{{ - 1}}{{50}} + \dfrac{1}{{25}} = \dfrac{1}{f} \\\
\implies \dfrac{{ - 1 + 2}}{{50}} = \dfrac{1}{f} \\\
\implies \dfrac{1}{f} = \dfrac{1}{{50}} \\\
In the final line of the above equation, we take reciprocal on both the sides and we get:
$f = 50\,{\text{cm}}$
The focal length of the lens is found out to be $50\,{\text{cm}}$ .
Again, we know,
1\,{\text{cm}} = {10^{ - 2}}\,{\text{m}} \\\
{\text{50}}\,{\text{cm}} = 0.50\,{\text{m}} \\\
The focal length in the S.I unit is $0.50\,{\text{m}}$ .

Again, the power is given by the following formula:
P = \dfrac{1}{f} \\\
\implies P = \dfrac{1}{{0.50\,{\text{m}}}} \\\
\implies P = + 2\,{\text{D}} \\\
Hence, the power of the lens is found out to be $+ 2\,{\text{D}}$ .

So, the correct answer is “Option B”.

Note:
While solving the numerical, remember that this is the case of a convex lens which treats far sightedness. The focal length will be positive as it is a converging lens, followed by the power of the lens which is also positive. Concave lens has negative power, as it is a diverging lens. The focal length must always be converted into an S.I unit before calculating power.