What Angle of Incidence Causes Total Reflection of X-rays in Calcite?

[Quelsita]

Problem:
The index of refraction for x-rays of wavelength 0.708A incident on calcite is 1-1.85E^-6. What angle of incidence will give total reflection of such X rays?


Using Bagg's Law: n(lambda)=2asin(theta), I get and angle of -45.07 degrees.
Is this all that is required of the problem...is it that simple?

Thanks.

 

[Redbelly98]

That's the wrong "n".

Bragg's law: n = any integer

But here n is the refractive index, not an integer. So Bragg's law doesn't apply.

What equation in optics deals with refractive index and angle of incidence? You might also look into "total internal reflection"

 

[Quelsita]

What equation in optics deals with refractive index and angle of incidence? You might also look into "total internal reflection"

OK, so with total internal reflection since we are looking for and angle that gives total reflection, so we want an angle greater than the critical angle, and since this means that (n1/n2) would be greater than 1, we cannot use Snell's Law: n₁sin(theta₁)=n₂sin(theata₂).

But if we let theata2=90, then we can solve for the critical angle:
theta_crit=arcsine(n₁/n₂)

-Is this correct?
-How does the wavelength of the X-ray apply?

 

[Redbelly98]

Your equation is correct.

The wavelength doesn't apply.

 

[Quelsita]

Ok. So is the second medium just air where n2=1?

 

[Redbelly98]

Ah, good question. Not sure what n is for x-rays in air. For optical wavelengths it's 1.0003.

But yes, the second medium is air.