(a) Consider a laser system consisting of an active medium placed between a pair of mirrors forming a resonator. Obtain an expression for the threshold population inversion required for the oscillations of laser. (20 marks)

(b) For a He – Ne laser s

Question

(a) Consider a laser system consisting of an active medium placed between a pair of mirrors forming a resonator. Obtain an expression for the threshold population inversion required for the oscillations of laser. (20 marks)

(b) For a He – Ne laser system, what will be the magnitude of Δω_D which represents FWHM of the line shape function g(ω), if resonant frequency ω₀ = 3 × 10¹⁵ s⁻¹ and temperature T = 300 K ? (10 marks)

(c) A cube of mass M and side 'a' is rotating with angular velocity ω around one of its edges, which is, say, along the x-axis. Obtain the expressions for its angular momentum and kinetic energy.
(Given that the I_XX = 2/3 Ma², I_YX = -1/4 Ma² and I_ZX = -1/4 Ma²) (20 marks)

UPSC Answer Check · Accepted Answer

(a)
Let the active medium have length L, mirror reflectivities R₁ and R₂, and distributed loss coefficient α. Let N₂ and N₁ be the upper- and lower-level populations with degeneracies g₂ and g₁. The net stimulated emission per unit volume is proportional to
N₂B₂₁ − N₁B₁₂ = B₂₁[N₂ − (g₂/g₁)N₁],
because the Einstein relation g₁B₁₂ = g₂B₂₁ gives B₁₂ = (g₂/g₁)B₂₁.
Hence the small-signal gain coefficient at angular frequency ω is
γ(ω) = σ(ω) ΔN, where ΔN = N₂ − (g₂/g₁)N₁,
and σ(ω) is the stimulated-emission cross-section. If g₁ = g₂, ΔN = N₂ − N₁.

For one round trip, the intensity becomes
I_after = I_before R₁R₂ exp[2(γ − α)L].
At threshold, steady oscillation begins, so I_after = I_before. Therefore
R₁R₂ exp[2(γ_th − α)L] = 1.
Solving,
γ_th = α + (1/(2L)) ln(1/(R₁R₂)).
Since γ_th = σ(ω) ΔN_th,
ΔN_th = [α + (1/(2L)) ln(1/(R₁R₂))]/σ(ω).
This is the threshold population inversion. If internal loss α = 0,
ΔN_th = ln(1/(R₁R₂))/(2Lσ(ω)).
If R₁ = R₂ = R, then
ΔN_th = −ln R/(Lσ(ω)).
The condition is small-signal, uniform gain, and steady-state cavity oscillation.

(b)
For Doppler broadening, an atom with velocity component v along the laser axis radiates at
ω = ω₀(1 + v/c).
Using the Maxwell speed distribution, the line shape is Gaussian. Its FWHM in angular frequency is found by setting the Gaussian equal to half its maximum:
exp[−m c²(ω − ω₀)²/(2kTω₀²)] = 1/2.
Thus
Δω_D = 2ω₀√(2kT ln2/(m c²)) = ω₀√(8kT ln2/(m c²)).
For the He–Ne laser, the lasing transition occurs in neon, so take
m_Ne ≈ 20.18 u = 20.18 × 1.66054 × 10⁻²⁷ kg = 3.351 × 10⁻²⁶ kg.
Given T = 300 K, k = 1.381 × 10⁻²³ J K⁻¹, c = 3.00 × 10⁸ m s⁻¹, ω₀ = 3 × 10¹⁵ s⁻¹.
kT = 1.381 × 10⁻²³ × 300 = 4.143 × 10⁻²¹ J.
m c² = 3.351 × 10⁻²⁶ × (3.00 × 10⁸)² = 3.016 × 10⁻⁹ J.
Therefore
√(8kT ln2/(m c²)) = √[8 × 0.6931 × 4.143 × 10⁻²¹/(3.016 × 10⁻⁹)]
= √(7.62 × 10⁻¹²) = 2.76 × 10⁻⁶.
Hence
Δω_D = 3 × 10¹⁵ × 2.76 × 10⁻⁶ = 8.28 × 10⁹ s⁻¹.
Δω_D ≈ 8.3 × 10⁹ s⁻¹, i.e. the ordinary-frequency FWHM is
Δν_D = Δω_D/(2π) ≈ 1.3 × 10⁹ Hz.

(c)
Take the origin at a corner on the edge along the x-axis, with body axes along the cube edges. The angular velocity vector is
Ω = ω i.
Using the rigid-body relation L = I · Ω,
Lₓ = Iₓₓ ω, Lᵧ = Iᵧₓ ω, L_z = I_zₓ ω.
Given Iₓₓ = (2/3)Ma², Iᵧₓ = −(1/4)Ma², I_zₓ = −(1/4)Ma²,
Lₓ = (2/3)Ma²ω,
Lᵧ = −(1/4)Ma²ω,
L_z = −(1/4)Ma²ω.
Thus
L = (2/3 Ma²ω)i − (1/4 Ma²ω)j − (1/4 Ma²ω)k.
Its magnitude is
|L| = Ma²ω√[(2/3)² + (−1/4)² + (−1/4)²]
= Ma²ω√(4/9 + 1/16 + 1/16)
= Ma²ω√(41/72).

The kinetic energy is
T = (1/2)Ω · L = (1/2)ωLₓ
= (1/2)ω[(2/3)Ma²ω]
= (1/3)Ma²ω².
The angular momentum is not along the rotation axis because the edge is not a principal axis of the cube.

(a) Consider a laser system consisting of an active medium placed between a pair of mirrors forming a resonator. Obtain an…

Model answer

What "Derive" is asking you to do

Structure that answers it

Where marks are lost

How this answer will be evaluated

Approach

Key points expected

Evaluation rubric

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Must cover

Loses marks

Earns more

Extra mark

Practice this exact question

More from Physics 2025 Paper I