Question

the person needs a lens of power -6 dioptres for correcting his distant vision .for correcting his near vision he need a lens of power plus two dioptre find the focal length of the lens and its nature required for his distance and near vision separately

Answer 1

For distant vision:
$p = \frac{1}{f}$
where 'p' is the power is dioptres and 'f' is the focal length in metres.

Therefore,
$- 6 = \frac{1}{f} \\ f = - \frac{1}{6} \\ f = - 0.1667m \\ f = - 16.667cm$
The nature of the lens is diverging.
The lens is concave as focal length of a concave lens is negative.


For Near vision:
$p = \frac{1}{f}$
$2 = \frac{1}{f } \\ f = 0.5m \\ f = 50cm$
The nature of the lens is converging.
The lens is convex as focal length of convex lens is positive.

Answer 2

Explanation:

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