First slide

Electromagnetic radiation - Dual nature of radiation and matter

Question

The eye can detect 5 x 10⁴ photons per square meter per sec of green light ( $λ = 5000 Å$ ) while the ear can detect 10^-13(W/m²). The factor by which the eye is more sensitive as a power detector than the ear is close to

Moderate

A

5
B

10
C

10⁶
D

15

Solution

$E = \frac{12375}{5000} = 2.475 eV \approx 4 \times 10^{- 19} J$
So the minimum intensity to which the eye can respond
$\begin{array}{l} I_{Eye} = (Photon flux) \times (Energy of a photon) \\ \Rightarrow I_{Eye} = (5 \times 10^{4}) \times (4 \times 10^{- 19}) ≃ 2 \times 10^{- 14} (W / m^{2}) \end{array}$
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.

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