37 (15 pts) Apply Snell’s law twice (external, then internal) to find it emerges at the
same angle.
38. (40 pts) Review the section on phase changes on reflection in Pedrotti section
23-3
Solution
(a) For the Fresnel rhomb shown, use the equations given to confirm that
53degrees will produce circular light if illuminated with light polarized 45 degrees to
the plane of incidence.
Light incident at 45 degrees will contribute equal TE and TM components to each
internal reflection. To make circular light, their needs to be a net phase difference
of 90 degrees, so a phase difference of 45 degrees is needed at each reflection:
 
 
sin 53   1 1.5
1 1.5 cos53 
sin 2 530  1 1.5
 
TE : tan  
cos 530
2
2
 
TM : tan  
2
2
2
0
2
0
TM  117.4 0
TE  72.30
  450
(b) Why is this device useful over a wider wavelength range than a wave plate that
relies on birefringence?
It’s ability to produce circular light is not dependent on matching the thickness to
specific multiple of the wavelength. Since it is based on reflections at a dielectric
interface, it will work about the same for all wavelengths, neglecting the small
amount of dispersion in glass at visible wavelengths.
(c) How could you produce elliptically polarized light with this Fresnel rhomb?
Rotate so that the incident polarization is not 45 degrees, or so that the angle of
incidence to the internal reflection is not 53 degrees, or put it in water to change the
relative index.
Grader: a is 20 pts, b and c are 10 pts.
39. (25 pts) Your research advisor has an excess of funding so you are asked to
make a polarizer out of a parallel slab of diamond (n = 2.44, assume no dichroism
or birefringence).
a) If your incident beam consists of unpolarized light of irradiance I0, what
irradiance of linearly polarized light can you make considering only a single external
reflection (ignore the reflection off of the back, which you will consider next)?
Draw a schematic of your polarizer and indicate the relevant angles.
68
n = 2.44
The polarizing angle for reflection will be at:
 p  tan 1 2.44  68
The reflection coefficient must be calculated at that angle from Fresnel’s
equations:
rTM  0
rTE  .7124
RTE  .51
Since half of the TE light will be reflected (which is half of the total TE plus
TM), the polarized light will be ¼Io.
b) If you now consider the first internal reflection from the back of the diamond
slab, what till be the total reflected irradiance?
68
n = 2.44
The TE light which enters the diamond will be partially internally reflected at
the bottom interface, then partially internally transmitted out of the top
interface, so the output polarized irradiance will be increased above ¼I o.
The initial transmission into the glass is:
tTE  0.287
And the reflection coefficient at the back interface is (taking into account the
new incident angle of 22.3 degrees and the internal reflection index of
0.4098):
r 'TE  0.713
Finally, the light must exit the front surface via an internal transmission:
t 'TE  1.713
So the net field that exits the front surface on the second reflection is:
tTE r 'TE t 'TE  0.35
The irradiance that escapes is therefore:
0.12 Io
Add this to the first reflection for the total power: 0.37 Io
c) Will multiple reflections from the front and back interfaces reduce the degree of
polarization of the reflected light?
The internal reflection and transmission will occur at 22.3 degrees according
to Snell’s law. At this angle, the internal r TM is also zero, so the TM mode
goes straight through. The externally reflected TE light remains perfectly
polarized.
Grader: I did not intend for them to consider interference between the front
and back beams - they would need to know the thickness.
Their answers may not exactly match mine due to rounding.
a) 10 pts, b) 10 pts, c) 5 pts.
40. (20 pts) Using Jones calculus, show that the effect of a Half Wave Plate on light
linearly polarized a inclination angle  is to rotate the plane of polarization through
an angle of 2. (The HWP may be used in this way as a “laser line rotator”,
allowing the plane of polarization of a laser beam to be rotated without having to
rotate the laser.
Solution:
cos  
Original light at angle : 

 sin  
1 0  cos    cos   cos  
Apply half wave plate: 




0  1  sin    sin    sin   
A change from  to – is a rotation of 2.
Grader: Multiplying by the Jones vector for a HWP is worth 12 pts,
recognizing the result is a 2 shift is worth 8 pts.
42. (10 pts) How thick should a half-wave plate of mica be in an application where
laser light of 633 nm is used? Appropriate refractive indices for mica are 1.599 and
1.594.
Solution:
To create a phase difference between the two polarization components, the
optical path difference must be /2:
n1t  n2t   2
t

2n1  n2 
= 63.3 m
Grader: 5 pts for setting up the first equation, 5 pts for the answer.