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A person having short-sight cannot see objects clearly beyond a distance of $1.5 m$ . What would be the nature and power of the corrective lens to restore proper vision?

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Convex lens;

- 0.67 D

Concave lens;

- 0.67 D

Convex lens;

+ 0.67 D

Concave lens;

+ 0.67 D

answer is B.

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Detailed Solution

Given, far point of defective eye

(u) = - \infty

Far point of normal eye

(v) = - 150 cm

Using lens formula,

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

\Rightarrow \frac{1}{f} = \frac{1}{- 150} - \frac{1}{(- \infty)}

\Rightarrow \frac{1}{f} = - \frac{1}{150} - 0

\Rightarrow f = - 150 cm

\Rightarrow f = - 1.5 m

Now, power of a lens is given by

P = \frac{1}{f}

\Rightarrow P = - \frac{1}{1.5}

\Rightarrow P = - 0.67 D

Therefore, the power of the lens to be used is

- 0.67 D

, and since the focal length of this lens is negative, it is a concave lens.

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