How Does Wave Interference Determine Intensity at a Point?

[Machodog]

Homework Statement

Two equal point sources S1 and S2 are a distance d apart. They emit waves in phase, of wavelength $$\lambda$$. P is a point on the line passing through the mid-point of S1S2 and making an angle $$\theta$$ with the centre line; the distance of P from S1 or S2 is very much greater than d.

Show that the intensity I of the disturbance at P is given by

$$I=2I_0\left\lbrace1+\cos \left(\frac{2\pi d\sin{\theta}}{\lambda}\right)\right\rbrace$$
Or
$$I=4I_0\cos^2 \left(\frac{\pi d\sin{\theta}}{\lambda}\right)$$

where $$I_0$$ is the intensity emitted by each source.

Homework Equations

Since we are told the distance of P from the two sources is very much greater than d, the source separation, we can assume that the lines S1P and S2P are parallel, in which case the path difference will be:

$$\frac{d \sin{\theta}}{\lambda}$$

And so the phase difference will be:

$$\frac{2\pi d\sin{\theta}}{\lambda}$$

The Attempt at a Solution

I've been trying to do this by adding two general waves with the same amplitude but out of phase like so:

$$A\sin{\omega t} + A\sin{(\omega t + \phi)}$$

Using a trig identity this becomes:

$$2A\sin{\left(\omega t + \frac{\phi}{2}\right)}cos{\frac{\phi}{2}}$$

Since intensity is amplitude squared, i just square this expression, but it doesn't really look like what I'm asked to show.I can do it another way, by using phasors and the cosine rule, but was just wondering if anybody could make the method above work, or explain to me why it doesn't work.

Any help appreciated and apologies for not having a picture.

 

[BigJDubs]

A:The second approach is the way to go.When two waves of equal amplitude emerge from two sources that are in phase, the superposition of the two waves at any point along their path will produce an intensity given by the formula$$I = I_0 ( 1 + \cos\phi )$$where $I_0$ is the amplitude of either wave and $\phi$ is the phase difference between the two at the chosen point. In this case the phase difference between the two waves at the point P is twice the phase difference between the waves at the mid-point of the two sources.You have correctly calculated the phase difference between the waves at the mid-point of the two sources to be $$\phi = \frac{2 \pi d \sin\theta}{\lambda}$$and consequently the phase difference between the two waves at P is $$2 \phi = \frac{2 \pi d \sin\theta}{\lambda}$$Substituting into the formula for the intensity gives$$I = I_0(1 + \cos 2 \phi)$$which can be rearranged as$$I = 2I_0 \left(1 + \cos \frac{2 \pi d \sin\theta}{\lambda}\right)$$or alternatively$$I = 4I_0 \cos^2 \left(\frac{\pi d \sin\theta}{\lambda}\right)$$

 

[thomate1]

The method you have attempted to use is not applicable in this situation. The equation you have derived is for the superposition of two waves with the same frequency and amplitude, but with a phase difference. In this problem, we are dealing with two point sources that are emitting waves in phase, meaning they have the same frequency, amplitude, and phase.

To solve this problem, we need to use the principle of superposition, which states that the total disturbance at a point is equal to the algebraic sum of the disturbances caused by individual sources. In this case, we have two point sources, S1 and S2, emitting waves of the same frequency and amplitude. The total intensity at point P will be the sum of the intensities caused by each source.

To find the intensity at point P, we need to consider the path difference between the waves from S1 and S2. As you correctly mentioned, the path difference will be:

\frac{d \sin{\theta}}{\lambda}

This path difference leads to a phase difference of:

\frac{2\pi d\sin{\theta}}{\lambda}

Since the two waves are in phase, this phase difference will lead to constructive interference at point P. This means that the total intensity at point P will be the sum of the individual intensities, which is equal to 2I_0. However, we also need to consider the angle of observation, \theta, which will affect the intensity at point P. This is where the cosine term comes in:

\cos \left(\frac{2\pi d\sin{\theta}}{\lambda}\right)

Multiplying this cosine term with the sum of the individual intensities, 2I_0, we get the final equation for the intensity at point P:

I=2I_0\left\lbrace1+\cos \left(\frac{2\pi d\sin{\theta}}{\lambda}\right)\right\rbrace

This is the same equation that we were asked to show. Alternatively, we can use the trigonometric identity:

\cos^2 \theta = \frac{1+\cos 2\theta}{2}

to rewrite the equation as:

I=4I_0\cos^2 \left(\frac{\pi d\sin{\theta}}{\lambda}\right)

which is the second equation that we were asked to show.

In summary, the method you have attempted to use