Optics - Dispersion Curve

The dispersion curve shown above relates the angular frequency to the wave number k. For waves with wave numbers lying in the range k₁ < k< k₂ which of the following is true of the phase velocity and the group velocity?

A. They are in opposite directions.
B. They are in the same direction and the phase velocity is larger.
C. They are in the same direction and the group velocity is larger.
D. The phase velocity is infinite and the group velocity is finite.
E. They are the same in direction and magnitude.

(GR9277 #79)

Solution:

Phase velocity: v_p = ω/k
Group velocity: v_g = dω/dk

In the range k₁ < k< k₂:
v_p → positive quantity
v_g → negative quantity (negative slope)
→ v_p and v_g are in opposite directions.

Answer: A

Sound and Wave - Group Velocity

The dispersion law for a certain type of wave motion is ω = (c²k²+m²)^½, where ω is the angular frequency, k is the magnitude of the propagation vector, c and m are constants. The group velocity of these waves approaches 

A. Infinity as k → 0 and zero as k → ∞
B. Infinity as k → 0 and c as k → ∞
C. c as k → 0 and zero as k → ∞
D. Zero as k → 0 and infinity as k → ∞
E. Zero as k → 0 and c as k → ∞

(GR8677 #59)

Solution:





As k → 0, v_g = 0
As k → ∞, v_g = c
since m ≪ →

Answer: E

Special Relativity - Index of Refraction

The measured index of refraction of X-rays in rock salt is less than one. This is consistent with the theory of relativity because

A. Relativity deals with light waves traveling in a vacuum only
B. X-rays cannot transmit signal
C. X-rays photons have imaginary mass
D. The theory of relativity predates the development of solid-state physics
E. The phase velocity and group velocity are different

(GR8677 #72)

Solution:

Index of refraction, n = ^c/_v

For X-rays, n  1 → v  c (violates the theory of special relativity)
However, this is OK, because in this case, v is phase velocity, v_p.

v_p does not represent the particle velocity.
The physical speed of particle is represented by group velocity, v_g.

Answer: E

Condensed Matter - Effective mass

Lattice forces affect the motion of electrons in a metallic crystal, so that the relationship between the energy E and wave number k is not the classical equation E = ħ²k²/2m, where m is the electron mass. Instead, it is possible to use an effective mass m* given by which of the following?

A.

B.

C.

D.

E.

(GR9277 #97) 

Solution: 

F = m*a

F = dp/dt = ħ dk/dt     (momentum, p = ħk)

a = dv_g/dt

Group velocity, v_g = dω/dk

Energy of electron, E = hf = ħω 
ω = E/ħ





F = m*a