# Microwaves and Interference problem

1. Dec 1, 2013

### srh38

I'm working on my physics prelab, but I can't seem to get this answer. The question states:

Measuring the detector´s output as a function of its position along the goniometer (look it up!), we see that there are maxima and minima in the signal. Starting at one of the maxima, we find 6 additional maxima after moving the reflector a distance of 8.91 cm. Therefore, the frequency of the microwaves from this generator is ____________.

The answer is supposed to be in GHz. I've tried using the equation nλ=dsinθ where x≈dsinθ and λ=c/f. That would give f = c*n/x where x = 8.91 cm and n = 6. This gave me 20.2 GHz, but apparently isn't the right answer... Help please??

2. Dec 1, 2013

### davenn

goniometer ?

did you spell that correctly ?
looking in wiki I cant see how it applies to Measuring the wavelength of an RF signal ??

Maybe it has a use that they or I am not familiar with

doing the math, 8.91cm /6 = 1.485cm
300 (c) / 1.485 = 20.2 GHz ( as you said)
desperately trying to remember if you would see maxima at 1/2 wavelength and as well as each wavelength

in that case 6 maxima would be 6 halfwaves
1.485 x 2 = λ = 2.97 therefore 300 / 2.97 = 10.1 GHz

hopefully someone else will confirm what the correct way is

Dave

Last edited: Dec 1, 2013
3. Dec 1, 2013

### srh38

@davenn I'm not sure how it applies either, I checked the spelling and it is correct. I don't think it affects how we calculate frequency though... I just don't understand what I'm doing wrong

4. Dec 1, 2013

### Simon Bridge

Welcome to PF;
$d\sin\theta$ would normally be a path difference.
Presumably the reflector controls the path difference somehow?

The existence of the interference pattern is shown by changing the angular position of the detector - which is the reference to the goniometer. The goniometer is then used to accurately fix the detector at the angle of one of the maxima.

Then you move a reflector (not the detector - leave the goniometer alone) some linear distance, the signal at the detector decreases, then increases ... when this has happened six times, the mirror is a distance x6 from it's initial position at x0. You are told that x6-x0=8.91cm

How does the position of the mirror affect the signal at the detector?

How much does the path diffreence have to change by in order to put a maxima, once more, on the detector angle?

Last edited: Dec 1, 2013
5. Dec 2, 2013

### sophiecentaur

Isn't a goniometer just an instrument for measuring angles? That would be what you wanted (although not a term which I have come across in this context) for measuring the directions of peaks and nulls in an interference (diffraction) pattern. ??? Sounds like shades of the Bragg diffraction formula, even.
So that we're all reading off the same hymn sheet, perhaps a diagram would help. The answer just has to be easy if we are solving the appropriate problem.

6. Dec 2, 2013

### Simon Bridge

@srh38: We don't know the setup - so we cannot tell how moving the "reflector" changes the path difference to the detector.
Presumably you know that - so we need to hear from you.
Till then - that's the best I can do.