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An eye has a near point distance of $0.75 m$ . What is the nature and power of the lens needed to reduce the near point distance to $0.25 m$ ?

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Convex lens;

- 2.67 D

Concave lens;

- 2.67 D

Convex lens;

+ 2.67 D

Concave lens;

+ 2.67 D

answer is C.

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Detailed Solution

Given, near point of normal eye

(u) = - 25 cm

Near point of defective eye

(v) = - 75 cm

Using lens formula,

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

\Rightarrow \frac{1}{f} = \frac{1}{- 75} - \frac{1}{(- 25)}

\Rightarrow \frac{1}{f} = \frac{50}{(75) (25)}

\Rightarrow f = 37.5 cm

\Rightarrow f = 0.375 m

Now, power of a lens is given by

\Rightarrow P = \frac{1}{f}

\Rightarrow P = \frac{1}{0.375}

\Rightarrow P = + 2.67 D

Therefore, the power of the lens to be used is

+ 2.67 D

and since the focal length of this lens is positive, it is a convex lens.

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