The minimum intensity of light to be detected by human eye is 10-10 W/m2. The number of photons of wavelength 5.6 x 10-7 m entering the eye, with pupil area 10-6 m2 ,per second for vision will be nearly
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a
100
b
200
c
300
d
400
answer is C.
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Detailed Solution
By using I=PA ; where P is radiation power, we get P=I×A ⇒nhctλ=IA ⇒nt=IAλhcHence , number of photons entering the eye per sec nt=10−10×10−6×5.6×10−76.6×10−34×3×108=282≈300