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# What will be the power of the eyeglasses worn by a person who is having a far point of $5m$?

Last updated date: 20th Jun 2024
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Hint: As per the question we can understand that the object can be considered to be at infinity. Therefore the image formed will be in the focus of the lens used. The power of a lens is described as the reciprocal of the focal length of the lens. The far point is the point of the image where the person can view a comfortable and clear image. This information will help you in solving the question.

First of all let us mention what all are given in the question. The far point of the eye glass used is given as,
$D=5m$
The far point is described as the point of the image formation which is provided with a clear and comfortable image. As per the question, the light rays are falling on the eye glasses from different regions. The object distance therefore can be considered as infinity. The image formed thus will be a virtual image.
Thus we can write that the object distance is given as,
$u=\infty$
And the image distance will be the focal length of the eye glasses used. It has been mentioned in the question that the image is formed at the far point. That is the image distance can be written as,
$v=-5m$
Applying this in the lens equation, we can write that,
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Substituting the values in the equation, we can write that,
$\dfrac{1}{f}=\dfrac{1}{-5}-\dfrac{1}{\infty }=\dfrac{1}{-5}-0=\dfrac{1}{-5}$
The power of the lens can be found by the equation,
$P=\dfrac{1}{f}$
That is,
$P=\dfrac{1}{f}=\dfrac{1}{-5} = -0.2D$

Note:
The focal length of an optical system is defined as a measure of the strength at which the system converges and diverges the light wave. It is defined as the inverse of the optical power of the system. The positive focal length represents that a system is converging the light. The negative focal length represents that the system will be diverging the same.