Search any question & find its solution
Question:
Answered & Verified by Expert
What minimum separation between two objects a human eye would be able to resolve, if the eye pupil diameter is $2 \mathrm{~mm}$ and the two objects are $20 \mathrm{~m}$ away from the eye?
(Assume, human eye to be equivalent to a convex lens and consider the average wave length of light as $600 \mathrm{~nm}$ )
Options:
(Assume, human eye to be equivalent to a convex lens and consider the average wave length of light as $600 \mathrm{~nm}$ )
Solution:
1448 Upvotes
Verified Answer
The correct answer is:
$7.32 \mathrm{~mm}$
Given, diameter of pupil, $d=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$ $D=20 \mathrm{~m}, \lambda=600 \mathrm{~nm}=6 \times 10^{-7} \mathrm{~m}$
For any aparture, limit of resolution is given as,
$\frac{y}{D} \geq \frac{1.22 \lambda}{d} \Rightarrow y \geq \frac{1.22 \lambda D}{d}$[For lens]
Putting the values
$y \geq \frac{6 \times 10^{-7} \times 20 \times 1.22}{2 \times 10^{-3}}$
$\Rightarrow \quad y \geq 6 \times 10^{-3} \times 1.22 \Rightarrow y \geq 7.32 \mathrm{~mm}$
Hence, minimum required separation is $7.32 \mathrm{~mm}$.
For any aparture, limit of resolution is given as,
$\frac{y}{D} \geq \frac{1.22 \lambda}{d} \Rightarrow y \geq \frac{1.22 \lambda D}{d}$[For lens]
Putting the values
$y \geq \frac{6 \times 10^{-7} \times 20 \times 1.22}{2 \times 10^{-3}}$
$\Rightarrow \quad y \geq 6 \times 10^{-3} \times 1.22 \Rightarrow y \geq 7.32 \mathrm{~mm}$
Hence, minimum required separation is $7.32 \mathrm{~mm}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.