Homework Help: Slit experiment maxima

1. Jul 27, 2011

Punkyc7

Monochromatic light illuminates a pair of thin parallel slits at normal incidence, producing an interference pattern on a distant screen. The width of each slit is 1/7 the center-to-center distance between the slits.

Which interference maxima are missing in the pattern on the screen?

every first will cancel
every 3rd will cancel
every 5th will cancel
every 7th will cancel
every 9th will cancel

Im leaning towards the seventh but why does the maxima just disappear? Shoulnt the maxima still be a maxima?

2. Jul 27, 2011

EinsteinKillr

Remember, that the diffraction pattern for a single slit gives us a series of minimas at quantum points in the pattern. This pattern does not go away when a second slit is introduced into the equation. The minimas produces by the single slit will in fact occur at the same angle as every so many maximums produced by the double slit diffraction pattern.

for maxima, $\delta$sin$\theta$=n $\lambda$

and for minima $\alpha$sin$\theta$=n$\lambda$

where $\alpha$=$\delta$/7

does that help?

3. Jul 27, 2011

Punkyc7

Ok but how does a maxima disappear, shouldnt they be evely spaced apart? Are we considering the distance between the slit where a single slit would be open?

4. Jul 27, 2011

PeterO

The maxima does not disappear - it was never there to start with.
OR
The maxima are missing, they have not disappeared.

The double slit interference comes about from the "interaction" of the two, overlapping, single slit patterns.

Think like this:

For the single slit pattern, the central bright band may be 5 cm wide on the screen. on each side is a 5 mm band of nothing, this is flanked by couple of bands 4 cm wide, then 5 mm bands of nothing etc etc [I am not interested here in actual widths of band - just the concept that there are bands].

With the second slit present, the interference pattern means that there should now be alternating bright and dark bands "evenly spaced" as you said. Suppose they are are each 2mm wide, with a 1 mm gap between them.

There will be 12-15 of them spread across the previous central broad band. the next one or two won't be there because they "should" have been in a position where the single slit diffraction patterns didn't produce any light anyway. Then comes another bunch of bands [across the secondary maxima] then one or two more missing, then some more in the tertiary maxima etc, etc.

This explanation was intended as a conceptual/pictorial description, not some analytic numerical specification.

Peter

5. Jul 27, 2011

EinsteinKillr

Consider the single slit experiment. You get a series of minima as a result, right? When you add a second slit, the single slit result doesn't go away. Rather it rules over the double slit phenomenon and makes it its slave. If you do the math, you'll find that the value of theta will coincide on a regular basis for both the minima produced by the size of the single slit and the maxima produced by the size of the double slit. Solve each equation for theta and set them equal. You will see why your intuition is correct.

6. Jul 27, 2011

Punkyc7

So what this question is asking is when will the minimum for the single slit the same as the maximum for the double slit. So that would mean that the it doesnt matter what wavelenght.

If that is correct than I think I understand it

7. Jul 27, 2011

PeterO

If the wavelength had been numerically important, you would have been given it. The fact that there was only one wavelength [monochromatic light] was vital.

Peter

8. Jul 27, 2011

Punkyc7

why is that?

9. Jul 27, 2011

PeterO

I re-read, and found it less unusual and a legitimate way of seeing the problem - thus my edit.

Peter

10. Jul 27, 2011

EinsteinKillr

What Petero is trying to say is that the wavelength does not matter in this particular solution. " It is not numerically important"
You are correct in thinking that this problem is true for all lambda.

11. Jul 27, 2011

Punkyc7

So if it is not depenedent on the wave lenght if you change the wave lenght would the missing maximum occur in the same spot. Im thinking it wouldnt but i want to be sure

12. Jul 28, 2011

PeterO

Correct, it should still be the nth maximum that is missing, it is just that the nth maximum would have been expected in a different place.

Peter