Question

Dan confirms the authenticity of an amethyst stone by a few tests. During one of the tests, he finds the angle of the refracted ray of light coming out of the amethyst into the air as 28°. If the refractive index of amethyst is 1.55, what is the incident angle?

Answer 1

Snell's law.
$n_{1}*sin \alpha _{1}= n_{2}*sin \alpha _{2}$
1=incident
2=refracted
Since it's coming out of the amethyst into air:
n1=1.55
n2=1.0003
$sin \alpha _{2}$ = $sin28$°

Substituting back into Snell's equation we get $sin \alpha _{1}=0.303$ so the angle is $\alpha _{1}=arcsin0.303=17.6$°

Answer 2

The answer is 47° I'm 100% positive