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Question: The minimum intensity of light to be detected by human eye is \(10^{- 10}W/m^{2}\). The number of ph...

The minimum intensity of light to be detected by human eye is 10−10W/m210^{- 10}W/m^{2}. The number of photons of wavelength 5.6×10−7m5.6 \times 10^{- 7}m entering the eye, with pupil area 10−6m210^{- 6}m^{2}, per second for vision will be nearly

A

100

B

200

C

300

D

400

Answer

300

Explanation

Solution

By using I=PA;I = \frac{P}{A}; where P = radiation power

⇒ P=I×AP = I \times A ⇒ nhctλ=IA\frac{nhc}{t\lambda} = IA ⇒ nt=IAλhc\frac{n}{t} = \frac{IA\lambda}{hc}

Hence number of photons entering per sec the eye

(nt)=10−10×10−6×5.6×10−76.6×10−34×3×108\left( \frac{n}{t} \right) = \frac{10^{- 10} \times 10^{- 6} \times 5.6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}= 300.