If the eye is kept very close to a converging lens (focal length = 10 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 :
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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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