Microwaves and Interference problem

[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??

 

[davenn]

goniometer ?

did you spell that correctly ?
looking in wiki I can't 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

 

[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

 

[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?

 

[sophiecentaur]

@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

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.

 

[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.