Worked example 6.4.1: Brewster windows of a He–Ne laser

A short calculation tied to Application 6.4.2. It uses only the Brewster angle and the S coefficient, and no new physics. A good way to use it: read the given quantities, produce the formulas from the application yourself, and only then put the numbers in.

Worked example 6.4.1

Brewster windows of a He–Ne laser

Given. He–Ne laser, $λ_{0} = 632.8 nm$ . The tube is closed by plane-parallel windows of fused silica, $n = 1.457$ , met by the beam at the Brewster angle. The mirrors are outside the tube.

Find. (a) $θ_{B}$ and $θ_{t}$ , (b) the angle at the inner face, (c) $∣ Γ_{⊥} ∣^{2}$ per surface and the S power left after a round trip (8 surfaces), (d) the P loss per round trip with a $1^{\circ}$ tilt error.

Try it yourself first.

Strategy

For nonmagnetic media $tan θ_{B} = n_{2} / n_{1}$ and $θ_{B} + θ_{t} = 9 0^{\circ}$ , Eq. (6.4.32). Each window has two faces and the beam crosses both windows twice per round trip: $2 \times 2 \times 2 = 8$ surfaces, each passing the fraction $1 - ∣Γ ∣^{2}$ of the power (Figure 6.4.25).

Worked example 6.4.1. A Brewster window seen from the side. The plate is tilted so that the beam meets it at \theta_B=55.5^\circ, runs inside at \theta_t=34.5^\circ to the normal, which is the Brewster angle of the inner face, and leaves parallel to its original direction. Only S light is reflected out of the beam. — **Figure 6.4.25.** Worked example 6.4.1. A Brewster window seen from the side. The plate is tilted so that the beam meets it at $θ_{B} = 55. 5^{\circ}$ , runs inside at $θ_{t} = 34. 5^{\circ}$ to the normal, which is the Brewster angle of the inner face, and leaves parallel to its original direction. Only S light is reflected out of the beam.

$θ_{B} = arctan 1.457 = 55. 5^{\circ}$ and $θ_{t} = 9 0^{\circ} - 55. 5^{\circ} = 34. 5^{\circ}$ .

The faces are parallel, so the beam meets the glass-air face at $34. 5^{\circ}$ , and $tan 34. 5^{\circ} = 0.687 = 1/1.457$ : this is the Brewster angle seen from the glass. Both faces are reflection-free for P.

With $η_{1} = η_{0}$ , $η_{2} = η_{0} / n$ (multiply numerator and denominator by $n / η_{0}$ ), $cos 55. 5^{\circ} = 0.5658$ , $cos 34. 5^{\circ} = 0.8245$ :

Γ_{⊥} ∣ Γ_{⊥} ∣^{2} = \frac{cos θ _{i} - n cos θ _{t}}{cos θ _{i} + n cos θ _{t}} = \frac{0.5658 - 1.457 \times 0.8245}{0.5658 + 1.457 \times 0.8245} = - 0.360, = 12.9 %.

The inner face gives $+ 0.360$ (the same interface seen from the other side). After a round trip the S wave keeps $(1 - 0.129)^{8} = 0.33$ of its power: it loses $67 %$ , the P wave nothing.

At $56. 5^{\circ}$ the outer face has $θ_{t} = 34. 9^{\circ}$ and $∣ Γ_{∥} ∣^{2} = 1.0 \times 1 0^{- 4}$ . The inner face, met at $34. 9^{\circ}$ , gives the same. Per round trip $1 - (1 - 1 0^{- 4})^{8} = 8 \times 1 0^{- 4} = 0.08 %$ .

Answer

θ_{B} = 55. 5^{\circ}, θ_{t} = 34. 5^{\circ}, ∣ Γ_{⊥} ∣^{2} = 12.9 %, S: - 67 % per round trip, P (1^{\circ} off): 0.08 %

Check

Windows at normal incidence would reflect $(\frac{0.457}{2.457})^{2} = 3.5 %$ per surface, $1 - (1 - 0.035)^{8} = 25 %$ per round trip, for both polarisations.

What the numbers say

The gain of a He–Ne tube is a few per cent per pass, far below the $67 %$ S loss, so only P oscillates and the output is linearly polarised in the plane of incidence. Normal-incidence windows would cost $25 %$ and the laser would not lase at all, while $1^{\circ}$ of mis-tilt costs under $0.1 %$ .

cf. Application 6.4.2