37 (15 pts) Apply Snell s law twice (external, then internal) to find it emerges at the same angle.

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1 37 (15 pts) Apply Snell s law twice (external, then internal) to find it emerges at the same angle. 38. (4 pts) Review the section on phase changes on reflection in Pedrotti section 3-3 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 and TM components to each internal reflection. To make circular light, their needs to be a net phase difference of 9 degrees, so a phase difference of 45 degrees is needed at each reflection: : tan sin TM: tan cos53 sin cos TM (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 pts, b and c are 1 pts.

2 39. (5 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 =.44, assume no dichroism or birefringence). a) If your incident beam consists of unpolarized light of irradiance I, 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 =.44 The polarizing angle for reflection will be at: p tan The reflection coefficient must be calculated at that angle from Fresnel s equations: r r TM R Since half of the light will be reflected (which is half of the total plus TM), the polarized light will be ¼I o.

3 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 =.44 The 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: t.87 And the reflection coefficient at the back interface is (taking into account the new incident angle of.3 degrees and the internal reflection index of.498): r '.713 Finally, the light must exit the front surface via an internal transmission: t ' So the net field that exits the front surface on the second reflection is: t r' t'.35 The irradiance that escapes is therefore:.1 I o Add this to the first reflection for the total power:.37 I o c) Will multiple reflections from the front and back interfaces reduce the degree of polarization of the reflected light?

4 The internal reflection and transmission will occur at.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 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) 1 pts, b) 1 pts, c) 5 pts. 4. ( 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. (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 Apply half wave plate: A change from to is a rotation of. cos cos cos 1 sin sin sin Grader: Multiplying by the Jones vector for a HWP is worth 1 pts, recognizing the result is a shift is worth 8 pts.

5 4. (1 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 and Solution: To create a phase difference between the two polarization components, the optical path difference must be /: n 1t nt t n n 1 = 63.3 m Grader: 5 pts for setting up the first equation, 5 pts for the answer.