First slide

Quantum Theory of Light

Question

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

Moderate

A

100
B

200
C

300
D

400

Solution

Intensity of light is given as $I = \frac{P}{A} = \frac{nE}{A}$ , when n is number of photons entering the eye, E is energy of each photons entering the eye, E is energy of each photon and A is area of pupil. $n = \frac{AI}{E} = \frac{ΛIλ}{hc}, where λ$ is wavelength of photon.
$n = \frac{10^{- 6} \times 10^{- 10} \times 5 \times 10^{- 7}}{6.34 \times 10^{- 34} \times 3 \times 10^{8}} = 300$
Hence, minimum 300 photons need to enter the eye of a person to detect light.

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