A person suffering from the eye-defect myopia (short-sightedness) can see clearly only up to a distance of 2 metres. What is the nature and power of lens required to rectify this defect?

A person suffering from the eye-defect myopia (short-sightedness), requires a concave lens to rectify this defect.

In order to find the power of the concave lens required, first, we have to calculate its focal length.

Given:

Far point of the myopic eye = 2 m. (the person can see an object kept at infinity if the image of the object is formed at the far point of 2 m from the eye).

Object distance $u$ = $\infty$

Image distance (or, far point in front of the lens) $v$ = $-$2 m

To find: Focal length, $f$ and power of the lens, $P$.

Solution:

From the lens formula we know that-

$\frac {1}{v}-\frac {1}{u}=\frac {1}{f}$

Substituting the given values we get-

$\frac {1}{(-2)}-\frac {1}{\infty}=\frac {1}{f}$

$-\frac {1}{2}-\frac {1}{\infty}=\frac {1}{f}$

$-\frac {1}{2}-0=\frac {1}{f}$    $(\because Any\ number\ divided\ by\ infinity\ is\ equa\l to\ 0)$

$-\frac {1}{2}=\frac {1}{f}$        $(by\ cross\ multiplication)$

$f=-2m$

Thus, the focal length, $f$ is -2m.

Now, 

We know that power of the lens is calculated as-

$P=\frac {1}{f(in\ meters)}$

Putting the value of $f$ in the formula we get-

$P=\frac {1}{-2}$

$P=-0.5D$

Thus, the power of the concave lens required to rectify the defect is -0.5 dioptres.

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Updated on: 10-Oct-2022

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