Interference Problem: Finding Maximum Radiation Intensity

[Saitama]

Homework Statement

A system illustrated in the figure consists of two coherent point sources 1 and 2 located in a certain plane so that their dipole moments are oriented at right angles to that plane. The sources are separated by a distance d, the radiation wavelength is equal to $\lambda$. Taking into account that the oscillations of source 2 lag in phase behind the oscillations of source 1 by $\phi$ ($\phi<\pi$), find the angles $\theta$ at which the radiation intensity is maximum.

Homework Equations

The Attempt at a Solution

Let the radiations from the sources $S_1$ and $S_2$ interfere at a point P. As the radiation from $S_1$ is already ahead by the given phase, I have
$$S_2P=S_1 P+d\cos\theta-\frac{\phi \lambda}{2\pi}$$
For constructive interference,
$$S_2P-S_1P=n\lambda$$
But solving the above equations doesn't give the right answer.

 

[ehild]

Homework Statement

A system illustrated in the figure consists of two coherent point sources 1 and 2 located in a certain plane so that their dipole moments are oriented at right angles to that plane. The sources are separated by a distance d, the radiation wavelength is equal to $\lambda$. Taking into account that the oscillations of source 2 lag in phase behind the oscillations of source 1 by $\phi$ ($\phi<\pi$), find the angles $\theta$ at which the radiation intensity is maximum.

Homework Equations

The Attempt at a Solution

Let the radiations from the sources $S_1$ and $S_2$ interfere at a point P. As the radiation from $S_1$ is already ahead by the given phase, I have
$$S_2P=S_1 P+d\cos\theta-\frac{\phi \lambda}{2\pi}$$

If you mean that S₂P and S₁P are distances of the point P from the sources, S₂P - S₁P≈dcosθ (when d<<S₁P and d<<S₂P). The phase difference between the rays arriving at P is (2π/λS₂P-Φ -2π/λS₁P), the effective path difference is S₂P-S₁P-Φλ/(2π) and that must be equal to integer times lambda.

ehild

 

[Saitama]

Hi ehild! :)

If you mean that S₂P and S₁P are distances of the point P from the sources, S₂P - S₁P≈dcosθ (when d<<S₁P and d<<S₂P). The phase difference between the rays arriving at P is (2π/λS₂P-Φ -2π/λS₁P), the effective path difference is S₂P-S₁P-Φλ/(2π) and that must be equal to integer times lambda.

ehild

Your method makes sense to me but solving it further doesn't seem to give the correct answer. Since $S_2P-S_1P=d\cos\theta$, for constructive interference, I get:
$$d\cos\theta-\frac{\phi \lambda}{2\pi}=n\lambda \Rightarrow \cos\theta=\left(n+\frac{\phi}{2\pi}\right)\frac{\lambda}{d}$$
But the given answer is:
$$\cos\theta=\left(n-\frac{\phi}{2\pi}\right)\frac{\lambda}{d}$$

 

[TSny]

It appears to me that the problem is worded such that dipole 2 lags dipole 1 in time. So, the phase lag in wave 2 is a phase lag in time, not distance.

 

[ehild]

It appears to me that the problem is worded such that dipole 2 lags dipole 1 in time. So, the phase lag in wave 2 is a phase lag in time, not distance.

You are right, the phase of the light from source 2 is (2pi/λ)s₂-(ωt-Φ) which results in a positive phase difference in distance. ehild

 

[Saitama]

It appears to me that the problem is worded such that dipole 2 lags dipole 1 in time. So, the phase lag in wave 2 is a phase lag in time, not distance.

Sorry if this is a stupid question but what is the difference between the two?

 

[ehild]

Read my previous post. -Φ is associated with the time term in the phase of the wave. You get "+Φ" added to the space part.


d\cos\theta+\frac{\phi \lambda}{2\pi}=n\lambda


ehild

 

[Saitama]

Read my previous post. -Φ is associated with the time term in the phase of the wave. You get "+Φ" added to the space part. d\cos\theta+\frac{\phi \lambda}{2\pi}=n\lambdaehild

I still don't seem to understand the difference between "phase lag in time" and "phase lag in distance". :(

Working backwards from the answer, the phase difference must be
$$\frac{2\pi}{\lambda}S_2P+\phi-\frac{2\pi}{\lambda}S_1P$$
But I don't get why we add $\phi$. What is the difference in answer if it is mentioned that "phase lag is in distance"?

 

[ehild]

A traveling wave is of the form E=Eosin(ωt-kr). If the oscillation of S2 lags behind S1, the time lag appears in the ωt term : it becomes ω(t-Δt). The phase changes by -ωΔt=-Φ, but you can incorporate that Φ into the -kr term, with positive sign.

ehild

 

[Saitama]

A traveling wave is of the form E=Eosin(ωt-kr). If the oscillation of S2 lags behind S1, the time lag appears in the ωt term : it becomes ω(t-Δt). The phase changes by -ωΔt=-Φ, but you can incorporate that Φ into the -kr term, with positive sign.

ehild

Sorry if this is going to sound stupid but in this case, would it be correct to say that oscillation of S2 lags by $\phi$ in time but is ahead by $\phi$ in distance?

Also, how do you conclude from the problem statement that the phase lag is in time not in distance?

 

[ehild]

Sorry if this is going to sound stupid but in this case, would it be correct to say that oscillation of S2 lags by $\phi$ in time but is ahead by $\phi$ in distance?

No, I do not think so. Lag means something that happens later. There is phase difference because of time lag, and because of longer distance travelled.

Also, how do you conclude from the problem statement that the phase lag is in time not in distance?

That was the problem statement:

Taking into account that the oscillations of source 2 lag in phase behind the oscillations of source 1 by ϕ...

The sources are oscillating dipoles. Their motion is not in phase.
The interference happens between the waves. Their phase difference is due to the time lag and the path difference.
ehild