Crystal Thickness for particular phase difference

[Sekonda]

1. Hey, the question I believe should be fairly simple to solve but it's been a while since I've looked over such content. There are two light waves that enter a crytstal at the same time, one linearly polarised parallel to the optic axis and one perpendicular to the optic axis, they have refractive indices η(1)=1.55 and η(2)=1.54. How thick must the crystal be for the light waves to exit with a 90 degree phase difference?



2. η=c/v



3. Right, so considering they have to have a phase difference of 90 degrees - that corresponds to a difference in quarter of a wavelength, so the amount of wavelengths in the thickness of the crystal must differ by just a quarter more i.e.

$$d=n\lambda_{1}=(n+\frac{1}{4})\lambda_{2}$$

Where 'd' is the thickness of the crystal, the number of wavelengths 'n' can be determined from

$$n=\frac{\frac{1}{4}\lambda_{2}}{\lambda_{1}-\lambda_{2}}=\frac{1}{4}\frac{\lambda_{2}}{\Delta \lambda}=\frac{1}{4}\frac{\eta _{1}}{\Delta\eta}$$

The last step I think we assume that the frequencies of the waves are the same... I can't remember how the last step is made unless we assume the frequencies of the waves are equal... Anyway, once 'n' is found we can substitute it into the first equation but I'm not sure how I find the final distance... I'll take another look at it.


Thanks SK

 

[Sekonda]

... We are told the wavelengths of the waves (in the vacuum) are both 589nm... Still, is this the correct way to approach the question?

I get an answer of about 14.7μm=d

 

[rude man]

... We are told the wavelengths of the waves (in the vacuum) are both 589nm... Still, is this the correct way to approach the question?

I get an answer of about 14.7μm=d

Me too.