# Light

## Contents

#### Problems on Light as a wave

Problem : Find an expression for the angular frequency of a wave in terms of the wavelength and phase velocity.

The most general form of a harmonic wave is given by ψ = A cos[k(x - vt)] , where v is the phase velocity and k is the wave number. Expanding this we have ψ = A cos(kx - kvt) . We know that the argument of the cosine must be dimensionless, so the expression kvt must be dimensionless, thus kv must be an inverse time, or the angular frequency of the wave (we know it is an angular frequency and not a regular frequency since we want the argument of the cosine to be in radians, which are dimensionless). Thus σ = kv . But the wavenumber is just k = 2Π/λ so σ = .

Problem : If the numbers in this problem are given in SI units, calculate the velocity of a wave given by the equation: ψ(y, t) = (9.3×104)sin[Π(9.7×106 y + 1.2×1015 t)] .

The speed is given by v = = = 1.24×108 meters per second. The direction is the along in the y -axis in the negative direction (since a minus sign causes the wave to advance to the right, and we have a plus sign here).

Problem : Write the equation for a wave with an amplitude 2.5×103 V/m, a period 4.4×10-15 seconds, and speed 3.0×108 m/s, which is propagating in the negative z -direction with value 2.5×103 V/m at t = 0 , z = 0 .

We want a wave of the form . The plus sign arises from the direction of travel: when t = 0 , z = 0 we have a peak at the origin, but as time increases ( z = 0 , t = Π/2 , for example) the peak advances to the left, and hence the wave is propagating in the negative direction as required. We can calculate σ , the angular frequency, from the period T = 1/ν = 2Π/σ . Thus σ = 2Π/T = = 1.43×1015 s -1 . We can compute k since we know that v = σk hence k = = = 4.76×106 m -1 . The amplitude is given and the cosine gives us the right phase (we could choose a sine an subtract a phase of Π/2 ). Thus:

Problem : Consider the wave ψ(x, t) = A cos(k(x + vt) + Π) . Find an expression (in terms of A) for the magnitude of the wave when x = 0 , t = T/2 , and x = 0 , t = 3T/4 .

When x = 0 we have ψ = A cos(kvt + Π) . At t = T/2 we then have ψ = A cos(kvT/2 + Π) . Now k = 2Π/λ , T = 1/ν and v = λν so kvT = 2Π . Thus we have ψ = A cos(2Π/2 + Π) = A cos(2Π) = A . In the latter case we have ψ = A cos(3×2Π/4 + Π) = A cos(5Π/2) = 0 .

Problem : Demonstrate explicitly that a harmonic function ψ(x, t) = A cos(kx - σt) satisfies the wave equation. What condition needs to be fulfilled?

Clearly the second (partial) derivatives with respect to y and z are zero. The second derivative with respect to x is:

 = - Ak 2cos(kx - σt)

The second derivative with respect to time is:

 = - Aσ 2cos(kx - σt)

Now the one-dimensional wave equation states that:

 =

From the derivatives calculated above this gives: - Ak 2cos(kx - σt) = . Canceling and rearranging this gives the required condition as: v = , which is just the result we stated for the phase velocity.