An eye has a near point distance of $0.75\,m$. What sort of lens in spectacles would be needed to reduce the near point distance to $0.25\,m$? Also calculate the power of the lens required. Is this eye long-sighted or short sighted?

Hint:A hypermetropic person with a near point of 0.75m will need converging spectacles to reduce the near point distance to 0.25m.We must first estimate the focal length in order to compute the power of the spectacles. After calculating this focal length we will find power.

Formula used:
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
Where $f$-focal length, $u$- object distance and $v$-image distance.
\[P = \dfrac{1}{f}\]
Where $P$=power and $f$= focal length.

Complete step by step answer:
To calculate the power of the spectacles we have to determine the focal length at first.
$u= -0.25\,m$ ( At normal near point distance of object).
$v=-0.75\,m$ (Defective eye's near point in front of lens).
We will now apply the formula:
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
\[\Rightarrow \dfrac{1}{f} = \dfrac{1}{{ - 0.75}} - \dfrac{1}{{ - 0.25}}\]
\[\Rightarrow \dfrac{1}{f} = \dfrac{1}{{0.25}} - \dfrac{1}{{0.75}}\]
\[\Rightarrow \dfrac{1}{f} = \dfrac{{6 - 2}}{{1.5}}\]
\[\Rightarrow \dfrac{1}{f} = \dfrac{4}{{1.5}}\]
We don’t need to solve further because, power has been asked and \[P = \dfrac{1}{f}\].Hence,
\[P = \dfrac{4}{{1.5}}\]
\[\therefore P = 2.66\,D\]
A positive power number implies long-sightedness, whereas a negative number suggests short-sightedness.. here as the power is positive, it is long-sightedness.

Note:Students usually get confused in the sign conventions. In this case remember to take both image and object distance negative. Also another very common mistake made is the application of formula. Many tend to apply \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\]. This is wrong. \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\] This is the mirror equation that has to be applied to mirror problems. In the lens formula we must apply \[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]. Another mistake could be that students apply \[P = \dfrac{{1 \times 100}}{f}\]to find power, this would be wrong. Here they have given all measurements in meters itself. The formula \[P = \dfrac{{1 \times 100}}{f}\]is applied only when you have to convert centimeters to meters.