The near-point of a person suffering from hypermetropia is at 50 cm from his eye. What is the nature and power of the lens needed to correct this defect? (Assume that the near-point of the normal eye is 25 cm).

A person suffering from the eye-defect hypermetropia (long-sightedness), requires a convex 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:

Near point of the myopic eye = 50 cm. (the person can see an object kept at the normal near point of 25 cm clearly if the image of the object is formed at the person's own near point of 50 cm from the eye).

Object distance $u$ = $-$25 cm

Image distance (or, near point of the defective eye) $v$ = $-$50 cm

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-

$\frac {1}{(-50)}-\frac {1}{(-25)}=\frac {1}{f}$

$-\frac {1}{50}+\frac {1}{25}=\frac {1}{f}$

$\frac {-1+2}{50}=\frac {1}{f}$    

$-\frac {1}{50}=\frac {1}{f}$        

$f=50cm=0.5m$

Thus, the focal length, $f$ is 0.5

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-

$P=\frac {1}{0.5}$

$P=\frac {10}{5}$

$P=+2D$

Thus, the power of the concave lens required to rectify the defect is +2 dioptres.

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Updated on: 10-Oct-2022

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