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If the eye is kept very close to a converging lens (focal length $=10 \mathrm{~cm}$ ) and at the optical centre of the lens and an object is kept at distance ' $d$ ' then the minimum distance ' $d$ ' of the object from the lens so that its image can be seen clearly by the defect free eye is: (A) $10 \mathrm{~cm}$

(B) $25 \mathrm{~cm}$

(C) $\frac{50}{3} \mathrm{~cm}$

(D) $\frac{50}{7} \mathrm{~cm}$

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Using, $\frac{1}{v}-\frac{1}{u}=\frac{1}{F} ; \Rightarrow \frac{1}{d}+\frac{1}{\left(-d^{\prime}\right)}=\frac{1}{F}$ (where $d$ ' is the image of the object from the lens, which behaves as the object for the eye)

$\Rightarrow \quad \frac{1}{d}=\frac{1}{F}+\frac{1}{d^{\prime}}$

For minimum $d, d$ ' should be minimum which is equal to $25 \mathrm{~cm}$ for eye.

Substituting ;

$\Rightarrow \quad \frac{1}{d}=\frac{1}{25}+\frac{1}{10}=\frac{10+25}{250}=\frac{7}{50}$

$\Rightarrow \quad d=\frac{50}{7} \text { Ans. }$

So, The correct option of this question is (D).

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