First slide

Quantum Theory of Light

Question

The eye can detect 5 x $10^{4}$ photons per square metre per sec of green light $(λ = 5000 \overset{o}{A})$ while the ear can detect $10^{- 13}$ $(W / m^{2})$ . The factor by which the eye is more sensitive as a power detector than the ear is close to

Difficult

A

5
B

10
C

$10^{6}$
D

15

Solution

$\begin{array}{l} E = \frac{12375}{5000} = 2 .475 eV \approx 4 \times 10^{- 19} J \\ So the minimum intensity to which the eye can respond \\ I_{Eye} = (photon flux) \times (Energy of a photon) \\ \Rightarrow I_{Eye} = (5 \times 10^{4}) \times (4 \times 10^{- 19}) = 2 \times 10^{- 14} (W / m^{2}) \\ Now as lesser the intensity required by a detector for detection, more sensitive it will be \\ \frac{S_{Eye}}{S_{Ear}} = \frac{I_{Ear}}{I_{Eye}} = \frac{10^{- 13}}{2 \times 10^{- 14}} = 5 i . e ., as intensity (power) \\ detector, the eye is five times more sensitive than ear . \end{array}$

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