Question

Least distance of distant vision of a long sighted man is $40cm$. He wishes to reduce it by $25cm$ by using a spectacle. What is the power and name the type of the lens used by him?

Answer 1

Hint: To solve this question, we need to use the lens formula. On substituting the values of the image and the object distances given in the question, we will get the focal length of the lens from which the power can be calculated. The sign of the focal length will tell the type of the lens used.

Formula used: The formulae used to solve this question are given by
1. $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$, here $f$ is the focal length of a lens, $v$ is the image distance, and $u$ is the object distance.
2. $P = \dfrac{1}{f}$, here $P$ is the power of a lens, and $f$ is its focal length.

Complete step-by-step solution:
Let us assume that the focal length of the lens in the spectacle used by the long sighted man to be equal to $f$.
Now, the near point of the man is given to be equal to $40cm$. Also it is given that he wishes to reduce it to $25cm$. So for an object kept at a distance of $25cm$ from the lens spectacle, the image must be formed at a distance of $40cm$ at the same side of the lens. So we have the object distance equal to $25cm$, and the image distance equal to $40cm$. From the lens formula we have
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
According to the Cartesian sign convention, both the object and the image distance must be negative. Therefore substituting $u = - 25cm$ and $v = - 40cm$ in the above equation, we get
 \[\dfrac{1}{f} = \dfrac{1}{{ - 40}} - \dfrac{1}{{ - 25}}\]
$ \Rightarrow \dfrac{1}{f} = \dfrac{1}{{25}} - \dfrac{1}{{40}}$
Taking the LCM, we have
$\dfrac{1}{f} = \dfrac{{40 - 25}}{{1000}}$
$ \Rightarrow \dfrac{1}{f} = \dfrac{{15}}{{1000}}$
Taking the reciprocal, we get
$f = \dfrac{{1000}}{{15}}cm$
$ \Rightarrow f = \dfrac{{10}}{{15}}m$
As we can see that the focal length of the lens is positive. So the lens must be convex.
Now, we know that the power of a lens is equal to the inverse of its focal length. So we get
$P = \dfrac{1}{f}$
$ \Rightarrow P = \dfrac{{15}}{{10}}D = 1.5D$
Hence, the lens used by the long sighted man is a convex lens having a power of $1.5D$.


Note: Do not forget to obtain the focal length of the lens in meters before calculating the power. Also, always check the sign of the focal length according to the type of lens needed for the eye disorder mentioned in the question. Since long sighted disorder is mentioned, so the lens must be convex which has a positive focal length.