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 maximum 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 maximum 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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Similarly as in the previous problem

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

For maximum 'd', d' should be maximum i.e. $d^{\prime}=\infty$

Substituting $; \frac{1}{d}=\frac{1}{F}+\frac{1}{\infty} \Rightarrow d=f=10 \mathrm{~cm}$.

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

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