Question

A professor reads a greeting card on his 50th birthday with a +2.5D glasses keeping the card 25cm away. 10 years later he reads the greeting card with the same glasses keeping the card 50cm away. What power should he wear now?
A. $2D$
B. $0.5D$
C. $2.25D$
D. $4.5D$

Answer 1

Hint: The professor must’ve experienced a change in eyesight, i.e., his near point must’ve changed as he aged. But he did not change the lens accordingly. First, we must determine the near point of his eye after 10 years, from the power of the lens given in the problem. Then, we can find the power of the lens needed to view the greeting from the distance of clear vision.

Formula used:
$P = \dfrac{1}{f}$
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Complete answer:
The formula for the power of lens is given by,
$P = \dfrac{1}{f}$
Here the power of the lens, ‘$P$’ is in Dioptre (D) and focal length ‘$f$’ is in meters.
It is mentioned in the question that, initially, the professor read a greeting card on the birthday with $ + 2.5D$ glasses. Let ‘${P_1}$’ be the initial power of the lens, which is given as $ + 2.5D$ and ‘${f_1}$’ be the corresponding focal length of the lens.
$\eqalign{
  & {P_1} = \dfrac{1}{{{f_1}}} \cr
  & \Rightarrow {f_1} = \dfrac{1}{{{P_1}}} = \dfrac{1}{{ + 2.5D}} = 0.4m \cr
  & \Rightarrow {f_1} = 40cm \cr} $
Initially the professor read the greeting card at $25cm$ length, which is the distance for clear vision. But, after 10 years his near point of the eye must’ve changed, as he aged. For this reason, he viewed the greeting from $50cm$.
If we consider $50cm$ as the object length ‘$u$' the corresponding near length is ‘$v$’, which can be found using lens formula as follows
$\eqalign{
  & \dfrac{1}{{{f_1}}} = \dfrac{1}{v} - \dfrac{1}{u} \Rightarrow \dfrac{1}{v} = \dfrac{1}{{{f_2}}} + \dfrac{1}{u} \cr
  & \Rightarrow \dfrac{1}{v} = \dfrac{1}{{40}} + \dfrac{1}{{50}} \cr
  & \Rightarrow v = \dfrac{{40 \times 50}}{{40 + 50}} = 200cm \cr} $
Now, the power of the lens must be changed to view the object at the minimum distance of clear vision, which is $25cm$. So, let the actual power of the lens needed be ‘${P_2}$’ and the corresponding focal length be ‘${f_2}$’. Applying the lens formula,
$\eqalign{
  & \dfrac{1}{{{f_2}}} = \dfrac{1}{v} - \dfrac{1}{u} \cr
  & \Rightarrow \dfrac{1}{{{f_2}}} = \dfrac{1}{{200}} - \dfrac{1}{{ - 25}} \cr
  & \Rightarrow {f_2} = \dfrac{{200 \times 25}}{{200 + 25}} = \dfrac{{200}}{9}cm \cr
  & \Rightarrow {f_2} = \dfrac{2}{9}m \cr} $
Now the corresponding power of the lens is
$\eqalign{
  & {P_2} = \dfrac{1}{{{f_2}}} = \dfrac{9}{2} = 4.5D \cr
  & \therefore {P_2} = 4.5D \cr} $

Therefore, the correct option is D.

Note:
The power of the lens, in Dioptre, is the reciprocal of focal length in meters. If the power of the lens is positive it indicates that the person is suffering from long-sightedness or hypermetropia. But if it is positive it indicates the person is suffering from near-sightedness or myopia.
The near point of the eye is the length from which the object can be viewed so that the object is focused on the retina i.e. when we view the object from a near point our eye doesn’t experience any kind of strain and we can see with ease.