Question

A double concave thin lens made out of glass $(\mu = 1.5)$ have radii of curvature $500$ cm. This lens is used to rectify the defect in vision of a person. The far point of the person will be at
A. $5m$
B. $2.5m$
C. $1.25m$
D. $1m$

Answer 1

Hint:Here,we are going to apply the concept of myopia and also know that the far point of a person must be at the total length of the lens used.

Formula used:
 $\dfrac{1}{f} = ({u^{ - 1}})\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$

Complete step by step answer:
We know that, for any lens, the image is formed at the feral length of the lens.
So, recently the vision of the person, the far point should be at the total length.
Calculate the total length, we will use the lens maker formula.
$\dfrac{1}{f} = (u - 1)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ . . . (1)
Where,
It is the total length.
$u$ is the refractive index of the lens.
We have,
$u = 1.5$
${R_1} = R_2^{} = 500un = 5m$
We changed it to meter because the fine answer in the options is in meters. Otherwise, we could solve it in the unit of centimeters as well.
By substituting the given values in equation (1), we get
$\dfrac{1}{f} = (1.5 - 1)\left( {\dfrac{1}{5} + \dfrac{1}{5}} \right)$
Substituting it, we get
$ = 0.5 \times \dfrac{2}{5}$
$ = \dfrac{5}{{10}} \times \dfrac{2}{5}$
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{5}$
By taking the reciprocal, we get
$t = 5m$
Since, the far point of equal to the total length, the far point should be at $5m.$
Therefore, from the above explanation the correct option (A). 5m.

Note: This is a simple question in which you just had to put values in the formula. But, their question becomes simple only if you know the concept that the far point should be at total length without knowing this concept, you would have not been able to guess which formula to use.