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Question

The minimum intensity of light to be detected by human eye is 1010W/m2. The number of photons of wavelength 5.6×107m entering the eye, with pupil area 106m2, per second for vision will be nearly

A
100
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B
200
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C
300
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D
400
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Solution

The correct option is C 300
By using I=PA; where P = radiation power
P=I×Anhctλ=IAnt=IAλhc Hence number of photons entering per sec the eye (nt)=1010×106×5.6×1076.6×1034×3×108=300.

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