Problem 4.4- Suppose n o and n e are given. In (a) you only need to find the magnitude of the group velocity. Problem #2 in HW 10 may be helpful. You can also directly use the definition of group velocity, i.e., v g = triangle k w (k), taking into account the equation of the wave normal surface.

4.4- Group Velocity and Phase Velocity

a.) Derive an expression for the group velocity of the extraordinary wave in a uniaxial crystal as a function of the polar angle 0 of the propagation vector.

b.) Derive an expression for the angle a between the phase velocity and the group velocity. This angle is also the angle between the field vectors E and D.

c.) Show that a = 0 when 0 = 0, ½ pi. Find the angle at which a is maximized and obtain an expression for a max. Calculate this angle a max for quartz with n o = 1.554, n e = 1.553.

d.) Show that for no or ne, the maximum angular separation a max occurs at 0 = 45 degrees; show that a max is proportional to [ n o – n e].

To solve problem 4.4, we are asked to derive various expressions related to the group velocity and phase velocity of an extraordinary wave in a uniaxial crystal. Let's go through each part step by step:

a) To find the magnitude of the group velocity, we can use the definition of group velocity: v_g = Δk / Δω, where Δk is the change in the wave vector and Δω is the change in the angular frequency. This equation can be written as v_g = dω / dk, where d represents the derivative.

In the problem, we are given that we need to find the group velocity as a function of the polar angle θ of the propagation vector. We can directly use the equation v_g = dω / dk, taking into account the equation of the wave normal surface.

To find the group velocity, we can start by using the definition of the extraordinary wave in a uniaxial crystal. The equation for the extraordinary wave is given by ω^2 / c^2 = k_e^2 - k_o^2, where ω is the angular frequency, c is the speed of light, and k_e and k_o are the extraordinary and ordinary wave vectors, respectively.

To get the group velocity, we need to differentiate this equation with respect to the wave vector k_e. Then, we solve for dω / dk_e. Finally, we can substitute the value of θ into the obtained expression to get the magnitude of the group velocity as a function of θ.

b) To find the angle α between the phase velocity and the group velocity, we can use the property that the wave vector k is in the direction of the phase velocity, and the group velocity is in the direction of the gradient of the dispersion relation. The angle α is also the angle between the field vectors E and D.

To find an expression for α, we can use the dot product of the wave vector k and the gradient of the dispersion relation to obtain cos α = (k · ∇)ω / |k||∇ω|. Now, we need to differentiate the dispersion relation with respect to k to find ∇ω. Finally, substitute the value of θ into the expression to obtain an equation for α.

c) To show that α = 0 when θ = 0, π/2, we need to substitute each of these values of θ into the equation derived in part b) and show that α becomes zero. To find the angle at which α is maximized, we can differentiate the expression derived in part b) with respect to θ and set it equal to zero. Solving this equation will give us the value of θ at which α is maximized. Then, substituting this value of θ into the expression for α will give us α_max.

To calculate α_max for quartz with n_o = 1.554 and n_e = 1.553, we can substitute these values into the expression for α_max derived in part c).

d) To show that the maximum angular separation α_max occurs at θ = 45 degrees, we can substitute θ = 45 degrees into the expression for α and show that it gives the maximum value. To show that α_max is proportional to (n_o - n_e), we can differentiate the expression for α_max with respect to the refractive index difference and show that it is proportional to (n_o - n_e).

By following these steps, you should be able to derive the expressions and solve the different parts of problem 4.4.