Question : Calculate the number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~10-10 W/m²). Take the area of the pupil to be about 0.4 cm² and average frequency of white light to be about 6×1014 Hz.
Similar Question : The minimum intensity of white light that our eyes can perceive is about 0.1 nW/m². Calculate the number of photons of this light entering our pupil (area 0.4 cm²) per second. (Take average wavelength of white light=500 nm and planck's constant=6.6×10^-34 Js)
Doubt by Jayant
Solution :
Intensity (I) = 10-10 W/m²
Area of cross section (A) = 0.4 cm²
= 0.4×10-4 m²
= 0.4×10-4 m²
ν = 6×1014 Hz.
we know,
I = P/A
I = P/A
P=IA
NE=IA [∵ P = NE]
N=IA/E
N=IA/hν
N = [10-10 ×0.4×10-4]/[6.6×10-34×6×1014]
N = [10-10 ×0.4×10-4]/[6.6×10-34×6×1014]
N = [0.4×10-14]/[39.6×10-20]
N = [4/396]×106
N = 0.0101×106
N = 1.01×104
N ≃ 104
N = [4/396]×106
N = 0.0101×106
N = 1.01×104
N ≃ 104
Hence the number of photons entering the pupil per second is 104.
