(a) Write conditions for working of a step-index optical fiber. In a step-index fiber, the core and cladding materials have refractive indices 1·50 and 1·43, respectively.

Find the following :

(i) Critical propagation angle

(ii) Acceptance angle

Question

(a) Write conditions for working of a step-index optical fiber. In a step-index fiber, the core and cladding materials have refractive indices 1·50 and 1·43, respectively.

Find the following :

(i) Critical propagation angle

(ii) Acceptance angle

(iii) Total time delay in 1 km length of the fiber

(iv) Total dispersion in 50 km length of the fiber

(b) Define streamline flow of a fluid. Using the equation of continuity for an isotropic fluid, find different components of total energy per unit volume.

(c) (i) What is the difference between Fresnel diffraction and Fraunhofer diffraction ?

(ii) What is resolving power of a telescope ? Why is the resolving power of microscope more with UV light than with visible light ?

UPSC Answer Check · Accepted Answer

(a) 
Conditions for working of a step-index optical fibre: the core refractive index must exceed the cladding refractive index, n₁ > n₂; light must enter the core within the acceptance cone; at the core–cladding boundary the angle of incidence must exceed the critical angle so that total internal reflection occurs; the boundary must be smooth and absorption/scattering small.
Given n₁ = 1.50, n₂ = 1.43.

(i) 
At the core–cladding interface, the critical angle is
θc = arcsin(n₂/n₁) = arcsin(1.43/1.50) = arcsin(0.953333) = 72.43°.
The maximum propagation angle with the fibre axis (critical propagation angle) is
θp = 90° − θc = 17.57°.
Equivalently, θp = arcsin(√(n₁² − n₂²)/n₁).
 Critical propagation angle = 17.57°; interface critical angle = 72.43°.

(ii) 
Numerical aperture:
NA = √(n₁² − n₂²) = √(1.50² − 1.43²) = √(2.25 − 2.0449) = √0.2051 = 0.4529.
Acceptance angle in air:
θa = arcsin(NA) = arcsin(0.4529) = 26.93°.
 θa = 26.93°.

(iii) 
Time for the axial ray over length L:
t_min = L n₁/c.
The extreme total-internal-reflection ray has path length L/cosθp and speed c/n₁, so
t_max = L n₁/(c cosθp).
Since cosθp = n₂/n₁,
t_max = L n₁²/(c n₂).
Therefore total time delay:
Δt = t_max − t_min = (L n₁/c)(n₁/n₂ − 1).
For L = 1 km = 1000 m, c = 3 × 10⁸ m/s:
L n₁/c = 1000 × 1.50/(3 × 10⁸) = 5.00 × 10⁻⁶ s = 5.00 μs.
n₁/n₂ − 1 = 1.50/1.43 − 1 = 0.07/1.43 = 0.048951.
So Δt = 5.00 μs × 0.048951 = 0.2448 μs = 2.448 × 10⁻⁷ s.
 Δt ≈ 0.245 μs.

(iv) 
Delay is proportional to length. For L = 50 km = 50000 m:
Δt = (50000 × 1.50/(3 × 10⁸)) × 0.048951
= 2.50 × 10⁻⁴ s × 0.048951
= 1.2238 × 10⁻⁵ s = 12.24 μs.
 Total dispersion in 50 km ≈ 12.24 μs.

(b) 
Streamline flow is a steady flow in which the velocity at any fixed point is independent of time. The streamlines are tangent to the velocity at every point, they do not cross, and the path of a fluid particle coincides with a streamline.

For an isotropic fluid, the equation of continuity is
∂ρ/∂t + ∇·(ρv) = 0.
For steady flow, ∂ρ/∂t = 0, so ∇·(ρv) = 0. For an incompressible isotropic fluid, ρ is constant, hence ∇·v = 0.
Using Euler’s equation for an inviscid isotropic fluid,
ρ(∂v/∂t + (v·∇)v) = −∇P − ρ∇Φ.
For steady flow and integration along a streamline,
P + 1/2 ρv² + ρΦ = constant.
Taking gravity as the body force, Φ = gh, so
E/V = P + 1/2 ρv² + ρgh.
Thus the total energy per unit volume has three components:
pressure energy per unit volume = P, in J/m³ or Pa;
kinetic energy per unit volume = 1/2 ρv²;
potential energy per unit volume = ρgh.
This holds for steady, inviscid, incompressible flow along a streamline.

(c)(i) 
In Fresnel diffraction, the source and/or screen are at finite distances from the aperture or obstacle. The incident and diffracted wavefronts are spherical or cylindrical, and the pattern is observed in the near field. No lenses are necessary, and the fringe pattern is generally irregular.
In Fraunhofer diffraction, the source and screen are effectively at infinity, so the wavefronts are plane. It is observed in the far field and is usually produced with lenses or a distant source. The pattern is regular with sharp maxima and minima and is treated by Fourier transforms.

(c)(ii) 
The resolving power of a telescope is its ability to form separate images of two nearby point objects. By the Rayleigh criterion, the minimum angular separation is
θ_min = 1.22λ/D,
so resolving power = 1/θ_min = D/(1.22λ).
Thus a larger aperture D or shorter wavelength λ gives better resolution.
For a microscope, resolving power = 1/d_min = 2n sinθ/λ = 2NA/λ.
Since resolving power is inversely proportional to wavelength, UV light has shorter λ than visible light, so it gives a smaller resolvable distance d_min and hence greater resolving power. Visible light has longer wavelength and therefore poorer resolution.

(a) Write conditions for working of a step-index optical fiber. In a step-index fiber, the core and cladding materials have…

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