Assignment 1

Q.1 : Light travelling in air is incident at an angle θ 0 = 33° to a glass plate, where θ 0 is measured between the incoming ray and the glass surface. Part of the light is reflected, and part is refracted. If the refracted and reflected rays make an angle of 90° to each other, what is the refractive index of the glass? What is the critical angle for this glass? Ans. : For Refractive index of the glass: By using Snell’s law, n 1 sinθ 1 = n 2 sinθ 2 Here, n 1 = n air = 1 sinθ 1 = sin ( 90 − θ ¿¿ 0 )= sin ( 90 − 33 ) = sin57 ᵒ ¿ n 2 = n glass sinθ 2 = sin 33 ᵒ ( θ 2 = θ 0 ¿ 1 ∗ sin57 ᵒ = n 2 sin33 ᵒ 1 ∗ 0.8386 = n 2 ( 0.5446 ) n 2 = 0.8386 0.5446 n 2 = 1.539 For the critical angle for this glass: The critical angle, which is determined by, is the angle at which light travelling through a media will collide with the border between that medium and another medium that has a lower refractive index without reflecting light from the first medium. ∅ c = sin − 1 ( n 1 n 2 ) ∅ c = sin − 1 ( 1 1.539 ) ∅ c = sin − 1 ( 0.6497 ) ∅ c = 40.51 ᵒ

Q.2 : A point source of light is 12 cm below the surface of a large body of still water ( 𝑛 = 1.33) with air above. What is the radius of the largest circle on the water surface from through which light can emerge? Ans. : Here, Snell's Law allows us to calculate the critical angle at which light passes from water to air, which we can then use to establish the radius of the greatest circle on the water's surface from which light may exit. In this case, the transition is from water ( n = 1.33 ) to air ( n = 1 ). Snell’s Law states this: n 1 sinθ 1 = n 2 sinθ 2 The source of light is directly below the surface of the water. So, θ 1 = 90 ᵒ and sinθ 1 = 1 n 1 = n air = 1 n 2 = n water = 1.33 1 ∗ 1 = 1.33 ∗ sin θ 2 θ 2 = sin − 1 ( 1 1.33 ) θ 2 = 48.75 ᵒ We will now use trigonometry to calculate the radius of the circle on the water's surface. The point source of light and the point on the circle where light emerges will form the circle's center, and the angle at their intersection will be θ 2 . Twelve centimetres separates the light source point from the light-emerging point on the circle. We can then find the radius by using the tangent: tan θ 2 = radius 12 cm radius = tan θ 2 ∗ 12 cm radius = tan 48.75 ∗ 12 cm Therefore, radius = 13.7 cm

E peak = √ P density 377 E peak = √ 11585.01 ∗ 10 6 377 E peak = √ 130.729 ∗ 10 6 E peak = 5543.4 V / m Q.4: (a) What is the meaning of “mode cut-off” and how can you determine it from the dispersion equation? Ans .: Mode cut-off: The opposite of saturation is the mode cut-off. When a transistor is in cutoff mode, it is off since there is no emitter current and no collector current. It appears to be almost entirely open. A transistor needs its base voltage to be lower than its emitter and collector voltages in order to enter cutoff mode. The number of modes that a fiber can sustain is known as its V-number, which is determined by multiplying the wavelength of light (l), the diameter of the core (d), and the fiber's numerical aperture (NA). The fiber is a single mode fiber (SMF) if V is less than 2.405. V greater than 2.405 indicates that the fiber is multimode (MMF). (b) Ans. : Matlab script and simulation plot are attached with this file. Please find attached files: “ Assignment1Q4b.m ” and “ Q4b_Modal field profiles in the transverse plane.fig ”. (c) Ans. : Matlab script and simulation plot are attached with this file. Please find attached files: “ Assignment1Q4c.m ” and “ Q4c_Modal field amplitude in the transverse plane.fig ”. (d) Give physical explanations for the observed behavior. Ans. : For Q. b:

Even Parity Mode:The modal field profile shows symmetric behaviour near the waveguide centre in the even parity mode. The reason for this symmetry is that in this mode, the electric field is even with respect to the waveguide centre. The field reaches its maximum in the centre of the core and decays exponentially towards the edges, with the propagation constant in the cladding controlling this declines. Odd Parity Mode: The modal field profile is antisymmetric about the waveguide centre in the odd parity mode. The reason for this antisymmetry is that the electric field in this mode has an odd parity in relation to the waveguide centre. Near the core's centre, the field is zero, and it reaches its greatest near the borders of the core, exhibiting an oscillatory behavior. For Q.c: Moderately Above Cut-off: Both even and odd parity modes show oscillatory behaviour with declining amplitude as they move away from the core-cladding interface at frequencies that are somewhat above cut-off. This is due to the fact that the modal fields are contained within the core, and the cladding region's mode leakage is represented by the exponential decay. Only Just above Cut-off: The modal field amplitude declines more quickly towards the cladding area at the cut-off frequency or just above it than it does at a significantly above cut-off. This quick decline indicates that the waveguide is the only thing supporting the mode. The field is primarily contained within the core, however it does extend somewhat into the cladding. Far Above Cut-off: The modal field amplitude shows little decay and is mostly contained inside the core at frequencies that are far above cut-off. This is due to the waveguide providing strong support for the mode and minimal field leakage into the cladding area. Stronger confinement of the mode is indicated by the more prominent oscillations inside the core. Q.5:

Q.6: Explain why 𝜈 must be an integer and, for a lossless fiber, 𝛽 must be real. Ans. : Concerning ν and β properties: An integer ν is required : The necessity for the eigenfunctions to be single-valued gives rise to this condition. The eigenfunctions rotate 2π and must return to their initial values; this forces ν to be quantized in order to maintain periodicity in θ . A lossless fiber requires that β be real : The solutions in a lossless system can't degrade or expand exponentially in space. Because of this, β must be real in order to prevent unlimited growth or decay from the exponential term exp(i β z).