(a) (i) Write down the system matrix for a combination of two thin lenses in paraxial approximation. Hence obtain the focal length of the combination and the positions of unit planes. (10 marks)

(ii) Consider a thin lens combination of two convex le

Question

(a) (i) Write down the system matrix for a combination of two thin lenses in paraxial approximation. Hence obtain the focal length of the combination and the positions of unit planes. (10 marks)

(ii) Consider a thin lens combination of two convex lenses of focal lengths f₁ = + 10 cm and f₂ = + 20 cm, respectively, separated by 25 cm. Determine the focal length of the combination and the positions of unit planes. (10 marks)

(b) The diameter of central zone of a zone plate is 2·4 mm. If a point source of light of wavelength 600 nm is placed at a distance of 5·0 m from the zone plate, calculate the position of the first image. (10 marks)

(c)

(i)

Consider three inertial frames of reference O, O' and O''. Let O' move with a velocity V with respect to O and O'' move with a velocity V' with respect to O'. Both velocities are in the same direction. Write down the transformation equations relating x, y, z, t with x', y', z', t' and also those relating x', y', z', t' with x'', y'', z'', t''. Hence obtain the relations between x, y, z, t and x'', y'', z'', t''. (The direction of velocity is chosen along the x-axis as per convention) 15 marks

(ii)

A galaxy in the constellation Ursa Major is receding from the Earth at 15000 km/s. If one of the characteristic wavelengths of light emitted by the galaxy is 550 nm, what is the corresponding wavelength measured by astronomers on the Earth? 5 marks

(ii) A galaxy in the constellation Ursa Major is receding from the Earth at 15000 km/s. If one of the characteristic wavelengths of light emitted by the galaxy is 550 nm, what is the corresponding wavelength measured by astronomers on the Earth? (5 marks)

UPSC Answer Check · Accepted Answer

(a)(i) 
In paraxial approximation, sin θ ≈ tan θ ≈ θ. Take a ray vector (y, θ), where y is height and θ is slope. The translation matrix through distance d is
T(d) = [[1, d], [0, 1]]
and the thin-lens matrix is
L(f) = [[1, 0], [-1/f, 1]].
For two thin lenses f₁ and f₂ separated by d, the system matrix from just before L₁ to just after L₂ is
M = L(f₂) T(d) L(f₁)
= [[1 - d/f₁, d], [-(1/f₁ + 1/f₂ - d/(f₁ f₂)), 1 - d/f₂]].
For any system matrix M = [[A, B], [C, D]], the effective focal length satisfies C = -1/f. Hence
1/f = 1/f₁ + 1/f₂ - d/(f₁ f₂).
For unit (principal) planes, h₁ = (D - 1)/C and h₂ = (1 - A)/C, measured from the first and second lenses respectively, positive to the right. Substituting A, C, D gives
h₁ = f d/f₂, h₂ = - f d/f₁.
Thus the first unit plane lies f d/f₂ to the right of L₁, and the second unit plane lies f d/f₁ to the left of L₂.

(a)(ii) 
Given f₁ = +10 cm, f₂ = +20 cm, d = 25 cm.
1/f = 1/10 + 1/20 - 25/(10×20)
= 0.10 + 0.05 - 0.125 = 0.025 cm⁻¹.
Therefore
f = 1/0.025 = 40 cm.
The first unit plane is
h₁ = f d/f₂ = (40×25)/20 = 50 cm
to the right of L₁. Since L₂ is 25 cm to the right of L₁, this is 25 cm to the right of L₂.
The second unit plane is
h₂ = - f d/f₁ = -(40×25)/10 = -100 cm
relative to L₂, i.e. 100 cm to the left of L₂, or 75 cm to the left of L₁.
 Final: f = +40 cm; H at +50 cm from L₁; H′ at -75 cm from L₁.

(b) 
The central-zone diameter is 2.4 mm, so its radius is
r₁ = 1.2 mm = 1.2×10⁻³ m.
Wavelength λ = 600 nm = 6.0×10⁻⁷ m.
For a source at distance u = 5.0 m and first image at distance v, the path difference between the axial ray and the ray through the first-zone edge is approximately
r₁²/2 (1/u + 1/v).
For the first image, this equals λ/2, so
r₁²(1/u + 1/v) = λ.
Thus
1/u + 1/v = λ/r₁² = 1/f, where f = r₁²/λ.
Now
f = (1.2×10⁻³)²/(6.0×10⁻⁷)
= 1.44×10⁻⁶/6.0×10⁻⁷ = 2.4 m.
Then
1/v = 1/f - 1/u = 1/2.4 - 1/5.0
= 5/12 - 1/5 = 13/60 m⁻¹.
Hence
v = 60/13 m = 4.615... m ≈ 4.62 m.
 Final: the first image is real and lies on the opposite side of the zone plate from the source at 60/13 m ≈ 4.62 m from the plate.

(c)(i) 
Let β = V/c and γ = 1/√(1 - β²). The Lorentz transformation from O to O′ is
x′ = γ(x - Vt), y′ = y, z′ = z,
t′ = γ(t - Vx/c²).
Let β′ = V′/c and γ′ = 1/√(1 - β′²). The transformation from O′ to O″ is
x″ = γ′(x′ - V′t′), y″ = y′, z″ = z′,
t″ = γ′(t′ - V′x′/c²).
Substituting the first pair into the second pair:
x″ = γ′[γ(x - Vt) - V′γ(t - Vx/c²)]
= γγ′[(1 + VV′/c²)x - (V+V′)t].
Similarly,
t″ = γ′[γ(t - Vx/c²) - V′γ(x - Vt)/c²]
= γγ′[(1 + VV′/c²)t - (V+V′)x/c²].
Define
u = (V+V′)/(1 + VV′/c²),
Γ = γγ′(1 + VV′/c²) = 1/√(1 - u²/c²).
Then
x″ = Γ(x - ut), y″ = y, z″ = z,
t″ = Γ(t - ux/c²).
This is again a Lorentz transformation with velocity u along the x-axis. Hence the velocity addition law is
u = (V+V′)/(1 + VV′/c²).

(c)(ii) 
For a receding source, the relativistic Doppler formula is
λ_obs = λ_emit √((1+β)/(1-β)),
where β = v/c.
Here
β = 15000/(3.00×10⁵) = 0.05.
Thus
λ_obs = 550 nm × √(1.05/0.95)
= 550 nm × √(21/19)
≈ 550 nm × 1.05132
= 578.22 nm.
 Final: λ_obs ≈ 578.2 nm.

(a) (i) Write down the system matrix for a combination of two thin lenses in paraxial approximation. Hence obtain the focal…

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