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Question: The near point of an elderly person is 50cm from the eye. Find the focal length and power of the cor...

The near point of an elderly person is 50cm from the eye. Find the focal length and power of the corrective lens that will correct his vision.

Explanation

Solution

Hint
The near point of an elderly person is 50cm. This means that the image forms at a minimum of 50cm. So if the object is kept at 25cm then we can find the focal length of the lens by
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} where the power of the lens will be given by the inverse of the focal length.
Formula used: In this question, we will use the formula,
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
Where ff is the focal length of the lens
uu is the object distance
and vv is the image distance.
The power of the lens is given by,
P=1fP = \dfrac{1}{f} where PP is the power of the lens.

Complete step by step answer
In the question, we are given that the near point of an elderly person is 50cm50cm. So when the object is placed at the minimum distance from where that eye can see without any strain given by 25cm, then the object needs to be formed at a distance of 50cm for the elderly person to see clearly.
So from the question, the image distance is given 50cm for the object distance minimum of 25cm.
Therefore since the image is virtual and is formed on the same side as that of the object so both the object and the image distance will be negative.
Hence,
u=25cmu = - 25cm and v=50cmv = - 50cm.
Now we can calculate the focal length of the lens by the formula,
1f=1v1u\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}
So by substituting the values of uu and vv we have,
1f=150+125\dfrac{1}{f} = - \dfrac{1}{{50}} + \dfrac{1}{{25}}
So by doing the subtraction,
1f=1+250\dfrac{1}{f} = \dfrac{{ - 1 + 2}}{{50}}
1f=150\dfrac{1}{f} = \dfrac{1}{{50}}
So we get the focal length of the lens required as f=50cm=0.5mf = 50cm = 0.5m
Now the power of the lens is given by, P=1fP = \dfrac{1}{f}
So by putting the value we get,
P=10.5D=2DP = \dfrac{1}{{0.5}}D = 2D. So the required power is P=2DP = 2D
Hence the focal length of the corrective lens required will be 0.5m and the power will be 2D.

Note
The near vision of normal adults is supposed to be 25cm but in elderly people this distance increases because the lens in the eye focuses the rays beyond the retina. This condition is termed as hypermetropia and can be corrected with the help of a suitable convex lens.