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 :
in Physics Edited by
3 views
0 Votes
0 Votes

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

in Physics Edited by
by
216 Points

1 Answer

0 Votes
0 Votes
 
Best Answer

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).

by
216 Points

Related Questions

Welcome To Informesia

Informesia Helps You To Prepare India's All States Board Exams (CBSE, ICSE, UP Board, BSEB, HPBOSE, RBSE, MSBSHSE) and Competitive Exam Like JEE (Main + Advanced), AIIMS, NEET, KVPY, NTSE, BITSAT, Olympiad, CLAT...etc.
For Any Query or Suggestion, Please Contact us on :
[email protected]

Connect us on Social Media