Question

Problem 1: Fibre optics (2 points) Figure 1: Sketches of the geometry of an optical fibre A long cylinder of dielectric material can guide light via total internal reflection. This idea dates back to the 1870s and forms the basis of the modern fibre optics communication technology. As long as the diameter of these fibres is large compared to the wavelength of the infalling light, its wave nature is negligible and we can use Geometric Optics to study the propagation of light. Optical fibres typically consist of a core, made from a material with nf, and cladding with a slightly smaller refractive index nc, which prevents light from leaking' out of the fibre due to frustrated total internal reflection. (a) (1pt) Consider the geometry illustrated on the left of Figure 1 to show that a light ray will be reflected L sin times inside a fibre of length L and diameter D, where 6, and θt are the initial incidence and trans- mission angles at the face of the fibre, respectively, and n,-1. Use this result to calculate how many reflections take place inside a fibre with L = 1m and D 2011m, when θί = 30 and nf = 1.55 (b) (1pt) As discussed during class, total internal reflection only takes place for internal reflection angles θ > e.. As shown in the right sketch of Figure 1, this implies that there will be a maximum value θmax of 6i at which the internal reflection angles is exactly 0c. All the rays hitting the face at angles greater than θmax will thus strike the interior walls at angles less than θ . In this case, no total reflection will take place and the intensity quickly leaks out of the fibre.max is consequently known as the acceptance angle, often expressed via the numerical aperture. This dimensionless number characterises the range of angles over which the system can accept light and is defined as Derive an expression for NA in terms of the refractive indices nc and nf and calculate NA and θтах for the case ni 1, n,-1.55 and nc-1.45

Answer 1

Ans:

1. Put values in the given equation, So Nr = 17040 or 17038 reflections take place
2. Applying snell’s law, and making right angle triangle at criticla angle part

So, ϴ₂ = π/2 - ϴ_c

sin ϴ₁ = n1 sin(π/2 - ϴ_c) = nf cos ϴ_c

cos ϴ_c = (1-sin² ϴ_c) ^1/2

sin ϴ₁ = n_f (1-sin² ϴ_c) ^1/2

We know,

Sin ϴ_c = n_c/n_f

Sin ϴ₁ = n_f( 1- (n_c/n_f)²)^1/2

NA = sin ϴ₁ = (n_f² – n_c²)^1/2

Here ϴ₁ = ϴ_max

NA = 0.547

ϴ_max = 33.16