Problem — Polariser and photon counting

Problem

A beam of red light ($\lambda = 650$ nm) with intensity $|\vv I| = 3\;\text{kW/m}^2$ strikes perpendicularly a polarising filter of area $S = 2\;\text{m}^2$. The outgoing intensity is one third of the incoming intensity. Calculate: (a) the angle $\alpha$ between the field $\vv E$ of the incident beam and the axis of the filter; (b) the number of photons per second passing through the filter.

Solution

Step 1 — Malus’s law. The intensity transmitted by a polariser is $I' = I\cos^2\alpha$, where $\alpha$ is the angle between the incident field and the axis. Imposing $I'/I = 1/3$:

$\cos^2\alpha = \frac{1}{3} \quad\Longrightarrow\quad \cos\alpha = \frac{1}{\sqrt{3}} \approx 0{,}577$

$\ev{\alpha = \arccos(0{,}577) \approx 54{,}7^\circ}$

Step 2 — Outgoing power. The outgoing intensity is $I' = I/3 = 1{,}0\;\text{kW/m}^2$, so

$P' = I'\cdot S = 1{,}0\cdot 10^{3}\cdot 2 = 2{,}0\cdot 10^{3}\;\text{W}$

Step 3 — Energy of a red photon.

$E_f = \frac{hc}{\lambda} = \frac{1240\;\text{eV·nm}}{650\;\text{nm}} \approx 1{,}91\;\text{eV} = 1{,}91\cdot 1{,}6\cdot 10^{-19} \approx 3{,}05\cdot 10^{-19}\;\text{J}$

Step 4 — Photons per second transmitted.

$\frac{N}{t} = \frac{P'}{E_f} = \frac{2{,}0\cdot 10^{3}}{3{,}05\cdot 10^{-19}} \approx 6{,}5\cdot 10^{21}\;\text{photons/s}$

$\ev{\alpha \approx 54{,}7^\circ, \qquad N/t \approx 6{,}5\cdot 10^{21}\;\text{photons/s}}$

Collegamenti

Argomenti: Quantum physics Concetti: Photon · Polarisation and Malus’s law Competenze: Unit conversion · Symbolic setup