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QJEEM15Ph90f8: JEE Main 2015 Physics - Class 12 JEEM15Ph90f8 Solution: Assuming human pupil to have a radius of and a comfortable…

Assuming human pupil to have a radius of 0.250.25 cmcm and a comfortable viewing distance of 2525 cmcm , the minimum separation between two objects that human eye can resolve at 500500 nmnm wavelength is :

Multiple Choice Options

  • A 100 μm100\,\mu m
  • B 300 μm300\,\mu m
  • C 1 μm1\,\mu m
  • D 30 μm30\,\mu m

Step-by-Step Verified Solution

sin⁡θ=0.2525=1100\sin \theta = {{0.25} \over {25}} = {1 \over {100}}

JEE Main 2015 (Offline) Physics - Wave Optics Question 129 English Explanation
Resolving power =1.22λ2μsin⁡θ=30 μm.= {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.

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