Problems on Refraction
Problem :
As light moves from air (n 1.00) to amber it deviates 18^{o} from its 45^{o} angle of
incidence. Which way does it bend? What is the speed of light in amber?
Light entering a denser medium refracts towards the normal. Thus the angle of refraction is
θ_{t} = 45  18 = 27^{o}. Using Snell's Law we have
n_{t} = 1.56.
The speed in amber is given by
v = c/n = 3.0×10^{8}/1.56 1.92×10^{8} m/s or
0.64c.
Problem :
A transparent fiber of index of refraction 1.6 is surrounded (cladded) by a less dense plastic of index
1.5. At what angle must a light ray in the fiber approach the interface so as to remain within the fiber?
This problem involves total internal reflection. The critical angle for staying within the fiber is given by:
sinθ_{c} = = 1.5/1.6 = 0.938. Thus
θ_{c} = 69.6^{o}. The ray must
approach the interface between the media at an angle of
69.6^{o} or greater to the normal.
Problem :
A light ray in air approaches a water surface (n 1.33) such that its electric vector is parallel to the
plane of incidence. If θ_{i} = 53.06^{o}, what is the relative amplitude of the reflected beam? What
about if the electric field is perpendicular to the plane of incidence?
We can apply the Fresnel Equations. In the first case we want the expression for
r_{  }. From
Snell's Law we can deduce that
sinθ_{t} = (n_{i}/n_{t})sinθ_{i} which implies
θ_{t} = 36.94^{o}. Then:
r_{  } = 0 

In the latter (perpendicular) case we have
r_{âä¥} = =  0.278 

In the former case, no light is reflected  this is called Brewster's angle as we shall see in the
section on polarization. For the perpendicular field the amplitude of the reflected wave is
0.278 as large as the incident wave. That is the reflected ray is about
(0.278)^{2} 0.08, or about
8% as bright as the incident ray (irradiance is proportional to the square of the amplitude).
Problem :
By what angle do blue light (λ_{b} = 460 nm) and red light (λ_{r} = 680 nm) disperse upon entering
(from vacuum) a medium with N = 7×10^{38}, ε = 1.94, and σ_{0} = 5.4×10^{15} Hz at
an incident angle of 20^{o} (the electron charge is 1.6×10^{19} Coulombs and its mass is 9.11×10^{31} kilograms)?
First we must calculate the index of refraction for both light frequencies. The angular frequency
of the blue light is
σ_{b} = 4.10×10^{15}Hz and for the red light
σ_{r} = 2.77×10^{15}.
Thus we have:
n_{r}^{2} = 1 + = 1 + = 1 + 0.472 

Thus
n_{r} = 1.213. Similarly for the blue:
n_{b}^{2} = 1 + = 1 + = 1 + 0.821 

Thus
n_{b} = 1.349. We can then calculate the angles of refraction of the two beams as they enter the
medium from Snell's Law. For the red:
1.213 sinθ_{r} = sinθ_{i}. This gives
θ_{r} = sin^{1}(sin(20^{o})/1.213) = 16.38^{o}. For the blue:
1.349 sinθ_{b} = sinθ_{i}.
Giving:
θ_{b} = 14.69^{o}. The difference between these two angles is
1.69^{o}, which is
the amount by which the different colored rays disperse.
Problem :
What is the rate of change of the speed of light with angular frequency in a dielectric medium?
We want
. But
v = c/n. So since
c is a constant, we can say
= c. Now:
Now we must take the derivative of this with respect to
σ treating everything else as a constant. This gives:
The required derivative is just
c times this: