First slide

Electromagnetic radiation - Dual nature of radiation and matter

Question

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

Moderate

A

100
B

200
C

300
D

400

Solution

$By using I = \frac{P}{A}; where P is radiation power, we get$
$P = I \times A \Rightarrow \frac{n h c}{t λ} = I A \Rightarrow \frac{n}{t} = \frac{I A λ}{h c}$
Hence , number of photons entering the eye per sec $(\frac{n}{t}) = \frac{10^{- 10} \times 10^{- 6} \times 5.6 \times 10^{- 7}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}} = 282 \approx 300$

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