(a) Consider a thick lens of thickness t made of a material of relative refractive index n. Let R₁ and R₂ be the radii of curvature of its two surfaces. Obtain the system matrix of the lens. (15 marks)

(b) Consider multiple reflections from a plane

Question

(a) Consider a thick lens of thickness t made of a material of relative refractive index n. Let R₁ and R₂ be the radii of curvature of its two surfaces. Obtain the system matrix of the lens. (15 marks)

(b) Consider multiple reflections from a plane parallel film of thickness h and refractive index n₂ and derive an expression for the total reflectivity from the surface of the film. (20 marks)

(c) A solid shaft of mass M, length l and radius r is to be replaced by a lighter hollow shaft of the same length l and having the same ratings of τ/θ, where τ is the couple and θ is the angle of twist. Estimate the percentage reduction in mass of the hollow shaft if the outer radius of the shaft is twice the inner radius. Assume the material of the new shaft is same as that of the replaced shaft. (15 marks)

UPSC Answer Check · Accepted Answer

(a)
Use the paraxial ray-transfer method. Take the ray vector as a column (y, nθ), where y is height and nθ is the reduced angle. Let R be positive if the centre of curvature lies to the right of the vertex. For refraction from index nᵢ to nₜ at a spherical surface of radius R, the matrix is
M_ref = [[1, 0], [-(nₜ - nᵢ)/R, 1]].
For translation through thickness t in medium n:
M_trans = [[1, t/n], [0, 1]].
At the first surface, nᵢ = 1, nₜ = n, radius R₁:
M₁ = [[1, 0], [-(n - 1)/R₁, 1]].
At the second surface, nᵢ = n, nₜ = 1, radius R₂:
M₂ = [[1, 0], [(n - 1)/R₂, 1]].
The system matrix is M = M₂ M_trans M₁. Multiplying,
M = [[A, B], [C, D]],
where
A = 1 - (n - 1)t/(n R₁),
B = t/n,
C = (n - 1)(1/R₂ - 1/R₁) - (n - 1)² t/(n R₁ R₂),
D = 1 + (n - 1)t/(n R₂).
Thus
M = [[1 - (n - 1)t/(n R₁), t/n], [(n - 1)(1/R₂ - 1/R₁) - (n - 1)² t/(n R₁ R₂), 1 + (n - 1)t/(n R₂)]].
For t → 0, C = (n - 1)(1/R₂ - 1/R₁), so 1/f = -C = (n - 1)(1/R₁ - 1/R₂). This is valid under the paraxial approximation.

(b)
Take a plane parallel film of refractive index n₂ and thickness h, with air on both sides. Let the amplitude reflection coefficient at an air-film interface be
r = (1 - n₂)/(1 + n₂).
At the lower film-air interface and at the upper film-air internal reflection, the reflection coefficient is -r. If t and t′ are the transmission coefficients, then t t′ = 1 - r².
Let θ₂ be the angle of refraction inside the film. The round-trip phase is
δ = 4π n₂ h cosθ₂/λ = 4π h √(n₂² - sin²θ₁)/λ,
where θ₁ is the external incidence angle. At normal incidence, δ = 4π n₂ h/λ.
The directly reflected amplitude is r. The first internally reflected beam has amplitude t t′(-r) exp(iδ). Each additional round trip multiplies the amplitude by r² exp(iδ). Hence
r_total = r + t t′(-r) exp(iδ) + t t′(-r) r² exp(i2δ) + ...
= r - (1 - r²) r exp(iδ)/(1 - r² exp(iδ)).
Simplifying,
r_total = r(1 - exp(iδ))/(1 - r² exp(iδ)).
Therefore the total reflectivity is
R = |r_total|²
= r² |1 - exp(iδ)|²/|1 - r² exp(iδ)|²
= 4 r² sin²(δ/2)/(1 + r⁴ - 2r² cosδ).
Writing R₀ = r² = ((n₂ - 1)/(n₂ + 1))²,
R = 4R₀ sin²(δ/2)/(1 + R₀² - 2R₀ cosδ).
If the two outer media are different, the general amplitude is
r_total = (r₁ + r₂ exp(iδ))/(1 + r₁ r₂ exp(iδ)),
and R = |r_total|². The result assumes no absorption, plane parallel surfaces, and coherence over the multiple reflections.

(c)
For torsion, the torsional rating is
τ/θ = G J/l,
where τ is the couple, θ is the angle of twist, G is the shear modulus, J is the polar second moment of area, and l is the length. Since the material and length are the same, equal τ/θ requires equal J.
For the solid shaft of radius r,
Jₛ = (π/2) r⁴.
For the hollow shaft, let the inner radius be a. The outer radius is 2a. Then
Jₕ = (π/2)((2a)⁴ - a⁴) = (π/2)(16a⁴ - a⁴) = (π/2)15a⁴.
Equating Jₛ and Jₕ:
(π/2)r⁴ = (π/2)15a⁴ ⇒ a⁴ = r⁴/15 ⇒ a² = r²/√15.
Mass of solid shaft:
M = ρπr²l.
Mass of hollow shaft:
Mₕ = ρπ((2a)² - a²)l = 3ρπa²l = 3ρπ(r²/√15)l = (3/√15)M.
So the fractional reduction is
1 - Mₕ/M = 1 - 3/√15 = 1 - √15/5.
Percentage reduction:
100(1 - 3/√15)% = 100(1 - √15/5)% ≈ 22.54%.
Thus the hollow shaft gives about 22.5% reduction in mass under the same torsional rating, same length, and same material.

(a) Consider a thick lens of thickness t made of a material of relative refractive index n. Let R₁ and R₂ be the radii of…

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