Question

A professor reads a greeting and received on his ${50^{th}}$ birthday with $+ 2.5{\text{D}}$ glasses keeping the card $25\;{\text{cm}}$ away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter $50\;{\text{cm}}$ away. What power of lens should he now use?

Answer 1

The lens power is defined as the reciprocal of the focal length. The power of the lens is measured in dioptres (D). There are positive focal lengths of converging (convex) lenses, so they also have positive power values.

Formula used: We will be using the following formula,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
$v$ is the Image distance
$u$ is the Object Distance
$f$ is the Focal Length

Complete Step-by-Step Solution:
A lens's power is the capacity to converge or diverge the light ray that falls on it. Lens power is characterized as the reciprocal focal length of the lens. A small focal length lens has a large converging or diverging power of a parallel beam of light.
On the ${50^{{\text{th }}}}$ birthday, he reads the card at a distance $25\;{\text{cm}}$ using a glass of $+ 2.5{\text{D}}$. Ten years later, his near point must have changed.
So after ten years,
$u = - 50\;{\text{cm}},f = \dfrac{1}{{2.5{\text{D}}}} = 0.4\;{\text{m}} = 40\;{\text{cm,}}v =$ near point
Now, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\quad$
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{u} + \dfrac{1}{f}$
Hence we can write,
$\Rightarrow \dfrac{1}{{ - 50}} + \dfrac{1}{{40}} = \dfrac{1}{{200}}$
So, near point $= {\text{v}} = 200\;{\text{cm}}$
To read the farewell letter at a distance of $25\;{\text{cm}}$,
$U = - 25\;{\text{cm}}$
For lens formula,
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Substituting the values we get,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{200}} - \dfrac{{ - 1}}{{ - 25}}$
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{200}} + \dfrac{1}{{25}} = \dfrac{9}{{200}}$
On calculating we get,
$\Rightarrow f = \dfrac{{200}}{9}\;{\text{cm}}$
$\Rightarrow f = \dfrac{2}{9}\;{\text{m}}$
$\Rightarrow$ Power of the lens $= \dfrac{1}{f} = \dfrac{9}{2} = 4.5{\text{D}}$
$\therefore$ He has to use a lens of power $+ 4.5{\text{D}}$.

Note:
The power of a lens combination would be the algebraic sum of the individual lens powers. There are negative focal lengths for diverging (concave) lenses, so they have negative power values as well. The lens power is specified as $P = \dfrac{1}{F}$ where the focal length is $F$. This is also referred to as the dioptre.