Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is:

1 μm

30 μm

100 μm

300 μm

Solution

30 μm

We can write resolving angle of necked eye as,

Where D is the diameter of the eye lens.
$fraction numerator straight Y over denominator 26 space straight X space 10 to the power of negative 2 end exponent end fraction space equals space fraction numerator 122 space straight X space 500 space straight X space 10 to the power of negative 9 end exponent over denominator 0.25 space straight X space 2 straight X space 10 to the power of negative 2 end exponent end fraction straight Y space equals space 30 space straight x space 10 to the power of negative 6 end exponent space straight m space equals space 30 space μm$

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Wave Optics

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is:

1 μm

30 μm

100 μm

300 μm

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For Daily Free Study Material Join wiredfaculty

Wave Optics

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is: 1 μm 30 μm 100 μm 300 μm

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is:

1 μm

30 μm

100 μm

300 μm