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 derive the expressions for the group velocity and the angle between the phase velocity and the group velocity, we can follow these steps:

a) Deriving the expression for the group velocity of the extraordinary wave:
1. Start with the definition of the phase velocity: v_p = w(k) / |k|, where w(k) is the angular frequency of the wave and |k| is the magnitude of the wave vector.
2. Taking the derivative of the phase velocity with respect to the wave vector k, we get: d(v_p)/d(k) = (d(w(k))/d(k) * |k| - w(k) * d(|k|)/d(k)) / |k|^2.
3. Assuming the wave vector k lies on the wave normal surface, we can write |k| = (n_o^2 * sin^2(Θ) + n_e^2 * cos^2(Θ))^0.5, where n_o and n_e are the refractive indices for ordinary and extraordinary waves, respectively, and Θ is the polar angle of the propagation vector.
4. Substitute the expression for |k| into the derivative from step 2.
5. Simplify and rearrange the terms to obtain an expression for the group velocity v_g in terms of n_o, n_e, Θ, and their derivatives with respect to k.

b) Deriving the expression for the angle between the phase velocity and the group velocity:
1. Start with the definition of the dot product between two vectors: A · B = |A| |B| cos(α), where α is the angle between the two vectors.
2. Consider two vectors: the phase velocity v_p and the group velocity v_g.
3. Express v_p and v_g in terms of their respective wave vectors k_p and k_g.
4. Write the dot product of v_p and v_g as a function of k_p, k_g, and their magnitudes.
5. Expand the dot product using the definition from step 1.
6. Use the expressions derived in part a) for v_p and v_g to express the dot product in terms of n_o, n_e, Θ, and their derivatives with respect to k.
7. Equate the expression for the dot product to the cosine of the angle a and solve for a.

c) To show that a = 0 when Θ = 0 or Θ = ½π, substitute these values into the expression derived in part b) for a. Verify that the angle a is zero.

To find the angle at which a is maximized and obtain an expression for a_max:
1. Take the derivative of the expression derived in part b) for a with respect to Θ.
2. Set the derivative equal to zero and solve for the angle Θ. This will give the angle at which a is maximized, denoted as Θ_max.
3. Substitute Θ_max into the expression derived in part b) for a to obtain a_max.

To calculate a_max for quartz with n_o = 1.554 and n_e = 1.553:
1. Substitute the given values of n_o and n_e into the expression derived in part b) for a_max.
2. Calculate the resulting angle a_max.

To derive expressions and solve the given problem, we will follow the steps mentioned in the instructions.

(a) To find the magnitude of the group velocity of the extraordinary wave in a uniaxial crystal as a function of the polar angle θ of the propagation vector, we can use the definition of group velocity:
v_g = (∂ω/∂k)

To find the expression for the group velocity, we need to use the equation of the wave normal surface, which can be obtained from Problem #2 in HW 10. Refer to that problem to get the equation of the wave normal surface.

(b) To derive an expression for the angle α between the phase velocity and the group velocity, we can use the relationship between the phase velocity and the group velocity:
cos(α) = v_p / v_g

You can derive this expression by writing down the expressions for the phase velocity and group velocity and then taking their dot product.

(c) To show that α = 0 when θ = 0, ½π, we can substitute these values of θ into the derived expression for α and show that it evaluates to 0. Similarly, to find the angle at which α is maximized, we can differentiate the expression for α with respect to θ and find the value of θ that maximizes α. The expression for α_max can be obtained by substituting this value of θ into the expression for α.

To calculate α_max for quartz with n_o = 1.554 and n_e = 1.553, you need to substitute these values into the derived expression for α_max.

(d) To show that the maximum angular separation α_max occurs at θ = 45 degrees, you can substitute θ = 45 degrees into the expression for α_max and show that it maximizes α_max. To show that α_max is proportional to (n_o - n_e), you can substitute the values of n_o and n_e into the expression for α_max and see if it can be written as α_max = A * (n_o - n_e), where A is a constant.

Remember to use the equations and relationships provided in the problem, as well as any relevant equations or concepts from your course material, to derive and solve for the desired expressions and values.