An interferometer with two arms is constructed above

[fobbz]

I have attached the problem as the image is needed for the question.

a) Basically, from what I understand the wave is being split, thus each chamber receives the total number of wavelengths/2. The chambers, however are vacuums, so the number of wavelengths passing in doesn't change right?
Looking at my formulae for interferometers for destructive and constructive interference I cannot see one that would contribute to solving this part of the problem?

my formulae: L=m$\lambda$/2 constructive interference, L=(m+.5)$\lambda$/2 for destructive interference.

b) I need a to find this part I believe.

c) Need b.

 

[haruspex]

the wave is being split, thus each chamber receives the total number of wavelengths/2.

It doesn't mean anything to say that each receives some number of wavelengths. Each receives half the total light.

The chambers, however are vacuums, so the number of wavelengths passing in doesn't change right?

Do you mean, the number of waves entering in a given interval of time doesn't change? That would be true. "Wavelengths" are lengths - they don't enter anything.

Looking at my formulae for interferometers for destructive and constructive interference I cannot see one that would contribute to solving this part of the problem?

5a is not a question about interference. It just asks how many wavelengths long would a 1cm evacuated tube be if the light has the given frequency. What formula do you know relating wavelength to frequency?

 

[fobbz]

well v=fλ. So if the wavelength is λ=c/f=(3E8 m/s) /(6E14 m) , λ=5E-7 m

So in 1 cm, 1 cm / 5E-7 = 20 000 wavelengths?

Okay that makes sense...

b) So next if the maxima changes to a minima, that means that an extra half wavelength is now being sent out making it destructive? So there would be 20 000.5 wavelengths?

 

[haruspex]

an extra half wavelength is now being sent out making it destructive? So there would be 20 000.5 wavelengths?

Wavelengths are not "sent out". Waves are sent out, and the number sent out depends on the frequency and the time period. What you mean is that the 1cm tube now represents 20000.5 wavelengths. (Or could it be 19999.5? How do you know which it is?)

 

[fobbz]

I know it is 20 000.5 because what changed was only the refractive index - the m order value did not change. So because it's now a minimum, (m+1/2) wavelengths would appear?

 

[haruspex]

I know it is 20 000.5 because what changed was only the refractive index - the m order value did not change. So because it's now a minimum, (m+1/2) wavelengths would appear?

You know it changed by half a wavelength. It is not instantly clear whether it increased or decreased by half a wavelength. But suppose it has increased. What is the refractive index?