# Homework Help: Interference from coherent light sources

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1. Jan 11, 2017

### CricK0es

1. The problem statement, all variables and given/known data

Four equally spaced coherent light sources (n=4) with a wavelength of 500nm are separated by a distance d = 0.1mm. The interference pattern is viewed on a screen at a distance of 1.4m. Find the positions of the principle interference maxima and compare their width with that for just two sources (n=2) with the same spacing.

2. Relevant equations

d . sinθ = mλ

3. The attempt at a solution

The positions of the principle interference maxima occur (for both) at an angle given by

θ = arcsin( mλ / d ) // So I can find these quite easily using:

m = 0, ±1, ±2 etc...
λ = 500 . 10^-9 m
d = 1 . 10^-4 m

However, I'm unsure how to find the widths of the maxima in question; for n=2 and n=4.

I would appreciate any guidance. Many thanks...

2. Jan 11, 2017

### TSny

Think about where the first minimum occurs for each case. Phasors might be helpful.

3. Jan 11, 2017

### CricK0es

I haven't done much on phasors yet, so I'm still not certain on how to use them.

The first minimum should occur when
d . sinθ = (m+1/2) λ

Solving for θ when m=0, should give me the angle of the first minimum. Can I then use these two angles (max and min), to find half the width of a principal maxima?

4. Jan 11, 2017

### TSny

This formula gives all the minima for two sources but only some of the minima for four sources. If you don't want to use phasors, then you might consider what would be the net intensity of all four sources when the angle is such that the first and third sources are out of phase.
Yes, if the angle θ is small. But, again, the formula you wrote will not necessarily give you the angle you need for 4 sources.

5. Jan 11, 2017

### CricK0es

Wait... I just found the equation sinθ = nλ / Nd, where N is the number of slits. Wouldn't setting N=4, and then solving for θ give me all the relevant minima?

In terms of total intensity, I've only got an equation stating it for two sources.

But I'm trying to think about the phasors you mentioned; given there are 4 sources, should the first minima occur when the phase difference is equal to π/2? (I'm trying to imagine a phasor diagram forming a square, so the pair cancel out: if that helps you follow my thought process.)

6. Jan 11, 2017

The formula $m \lambda=d sin(\theta)$ gives the position of the primary maxima for any number of equally spaced sources, essentially for number of sources $N$ from 2 to 30,000 or more, as is found on a diffraction grating. $m$ can of course be any integer, positive, negative, or zero. (Meanwhile, your formula in post #5 is incorrect. Also for $N$ sources, (where $N>2$) you really don't need to find zeros or minima.)

7. Jan 11, 2017

### CricK0es

Yeah I agree with that. It's just finding the minima for N slits. Would sinθ = nλ / Nd be appropriate for this?

8. Jan 11, 2017

### TSny

This formula certainly doesn't work in general. Try N = 2 with n = 2.

Yes, very good!

9. Jan 11, 2017

No. If you want the general formula for the intensity $I$ for $N$ equally spaced sources, it is $I(\theta)=I_o \frac{sin^2(\frac{N \phi}{2})}{sin^2(\frac{\phi}{2})}$ where $\phi=\frac{2 \pi}{\lambda} d \, sin(\theta)$ and the zeros occur when the numerator is zero provided the denominator is also not zero. When both numerator and denominator are zero, this is where primary maxima occur, and the limit near these points is $I(\theta_{max})=N^2 I_o$. You will notice if you set the denominator equal to zero, that you get $m \lambda=d sin(\theta)$ . Also, in general, for large $N$, the zeros aren't of much interest other than to note that there will be secondary maxima somewhere between each pair of zeros. For large $N$, the primary interest is in the location of the primary maxima. $\\$ Additional editing.. I see by looking at the numerator of the formula I provided for $I (\theta)$, you can get the formula you are using (for the location of the zeros), but only provided it is not the location of a primary maximum, where the denominator is also zero.

Last edited: Jan 11, 2017
10. Jan 11, 2017

### CricK0es

Sorry guys, I'm completely lost. I'm flicking through my book seeing how to apply these things, but it's just not clicking. I'll sleep on it and see if anything springs to mind in the morning. Thank you so far though...

Also, Charles, I misread your post (post 6). That's why my response may have sounded odd, apologies

11. Jan 11, 2017

My post #9 is much more thorough than my post #6 and should provide you with the info you need. $\\$ One other thing worth mentioning is what causes primary maxima for equally spaced sources. If two adjacent sources completely constructively interfere, than any other sources that are equally spaced will also all constructively interfere, because adjacent ones have complete constructive interference. e.g. if 1 is constructive with 2, and 2 is constructive with 3, then 1 and 3 must constructively interfere, etc. The condition for this is $m \lambda =d sin(\theta)$, i.e. the path difference (to a location in the far field) between adjacent sources must be an integer number of wavelengths. This is why the condition for primary maxima is the same regardless of the number of sources $N$. $\\$ Editing... Post #9 might even be overly thorough, and perhaps this last explanation might prove to be quite useful.

Last edited: Jan 11, 2017
12. Jan 12, 2017

### BvU

13. Jan 12, 2017

### TSny

Yes. So, at what angle θ is the phase difference between neighboring sources equal to π/2?

14. Jan 12, 2017

### CricK0es

When θ = arcsin( π / 2d )?

15. Jan 12, 2017

Deleted comment. It needed a correction. Post #11 tells most of what you need. Additional comment: For $N$ sources, you will find there are $N-1$ zeros in intensity between the primary maxima, and thereby $N-2$ secondary maxima. For large $N$ the most important feature is the primary maxima, but it is a good exercise to work with the phasor diagrams, etc., especially for small $N$ to see what the phasor diagram tells you. $\\$ You can also evaluate the numerator of post #9 to find the zeros for the numerator, and use them in the phasor diagram (to get zero for the resultant), provided it doesn't also make the denominator zero. In the case of the denominator equal to zero, (and your phasor diagram would also give this result), you will have a primary maximum with constructive interference between all of the sources.

Last edited: Jan 12, 2017
16. Jan 12, 2017

### TSny

This is not quite it. Recall how to express the path difference of the rays from two neighboring sources in terms of d and θ. What should this path difference be equal to in order to get a phase difference of π/2? Hint: What would be the phase difference if the path difference is λ?

17. Jan 12, 2017

### CricK0es

A quarter of the wavelength? If the path difference is λ, then surely the phase difference is just 2π/0

18. Jan 12, 2017

### TSny

Yes. So now what expression do you get for θ if you want the first minimum of 4 sources?

19. Jan 12, 2017

### CricK0es

Would it just be θ = arcsin( λ(n + 1/4) / d ) ? Sorry about being an ignoramus with this

20. Jan 12, 2017

### TSny

How do you express the path difference in terms of d and θ?

Once you have this expression, set it equal to a quarter of a wavelength.

21. Jan 12, 2017

### CricK0es

Sinθ = (path difference) / d ∴ θ = arcsin( λ / 4d )

22. Jan 12, 2017

### TSny

Yes. That's it. You mentioned earlier how to get the width of the central maximum from θ. You can then compare to 2 slits.

23. Jan 12, 2017

### CricK0es

Okay I see. I'm sorry I took so long, so thank you for being patient ;)

24. Jan 12, 2017

### TSny

You should get that there is a simple relation between the widths of the maxima for 4 slits as compared to 2 slits (assuming small angle approximation: sinθ ≈ θ).

No problem.