Students preparing for an exam will want to understand the following questions and prob-lems thoroughly enough to be able to work them without referring back to previous chapters.
True and False Questions
R1 T or F: The optical index of any material (not vacuum) varies with frequency.
R2 T or F: The frequency of light can change as it enters a crystal (consider low intensity—no nonlinear effects).
R3 T or F: The entire expression E0ei(k·r−ωt) associated with a light field (both the real part and the imaginary parts) is physically relevant.
R4 T or F: The real part of the refractive index cannot be less than one.
R5 T or F: s-polarized light and p-polarized light experience the same phase shift upon reflection from a material with complex index.
R6 T or F: When light is incident upon a material interface at Brewster’s angle, only one polarization can transmit.
R7 T or F: When light is incident upon a material interface at Brewster’s angle one of the polarizations stimulates dipoles in the material to oscillate with orientation along the direction of the reflected k-vector.
R8 T or F: The critical angle for total internal reflection exists on both sides of a material interface.
R9 T or F: From any given location above a (smooth flat) surface of water, it is possible to see objects positioned anywhere under the water.
R10 T or F: From any given location beneath a (smooth flat) surface of water, it is possible to see objects positioned anywhere above the water.
R11 T or F: An evanescent wave travels parallel to the surface interface on the trans-mitted side.
R12 T or F: When p-polarized light enters a material at Brewster’s angle, the intensity of the transmitted beam is the same as the intensity of the incident beam.
R13 T or F: For incident angles beyond the critical angle for total internal reflection, the Fresnel coefficients ts and tp are both zero.
R14 T or F: As light enters a crystal, the Poynting vector always obeys Snell’s law.
R15 T or F: As light enters a crystal, the k-vector does not obey Snell’s for the extraordinary wave.
Problems
R16 (a) Write down Maxwell’s equations.
(b) Derive the wave equation for E under the assumptions that Jfree = 0 and P = 0χE. Note: ∇ × (∇ × f ) = ∇ (∇ · f ) − ∇2f .
(c) Show by direct substitution that E (r, t) = E0ei(k·r−ωt) is a solution to the wave equation. Find the resulting connection between k and ω. Give appropriate definitions for c and n, assuming that χ is real.
(d) If k = kˆz and E0= E0x, find the associated B-field.ˆ
(e) The Poynting vector is S = E × B/µ0, where the fields are real. Derive an expression for I ≡ hSit.
R17 A horizontal and a vertical polarizer are placed in series, and horizontally polarized light with Jones vector
1 0
enters the system.
Figure 5.7
(a) What is the Jones vector of the transmitted field?
(b) Now a polarizer at 45◦ is inserted between the other two polarizers. What is the Jones vector of the transmitted field? How does the final intensity compare to initial intensity?
125 (c) Now a quarter wave plate with a fast-axis angle of 45◦ is inserted between the two polarizers (instead of the polarizer of part (b)). What is the Jones vector of the transmitted field? How does the final intensity compare to initial intensity?
R18 (a) Find the Jones matrix for half wave plate with its fast axis making an arbitrary angle θ with the x-axis.
HINT: Project an arbitrary polarization with Ex and Ey onto the fast and slow axes of the wave plate. Shift the slow axis phase by π, and then project the field components back onto the horizontal and vertical axes. The answer is
cos2θ − sin2θ 2 sin θ cos θ 2 sin θ cos θ sin2θ − cos2θ
(b) We desire to attenuate continuously a polarized laser beam using a half wave plate and a polarizer aligned to the initial polarization of the beam (see figure).
The fast axis of the half wave plate is initially aligned in the direction of polar-ization and then rotated through an angle θ. What is the ratio of the intensity exiting the polarizer to the incoming intensity as a function of θ?
Figure 5.8 Polarizing Elements
R19 Consider an interface between two isotropic media where the incident field is defined by
Ei=Ei(p)(ˆy cos θi− ˆz sin θi) + ˆxEi(s) ei[ki(y sin θi+z cos θi)−ωit]
The plane of incidence is shown in Fig. 5.9
(a) By inspection of the figure, write down similar expressions for the reflected and transmitted fields (i.e. Er and Et).
(b) Find an expression relating Ei, Er, and Et using the boundary condition at the interface. From this expression obtain the law of reflection and Snell’s law.
(c) The boundary condition requiring that the tangential component of B must be continuous leads to
ni(Ei(p)− Er(p)) = ntEt(p) ni(Ei(s)− Er(s)) cos θi= ntEt(s)cos θt Use this and the results from part (b) to derive
rp ≡ Er(p)
Ei(p) = −tan (θi− θt) tan (θi+ θt) You may use the identity
sin θicos θi− sin θtcos θt
sin θicos θi+ sin θtcos θt = tan (θi− θt) tan (θi+ θt)
z-axis x-axis
directed into page
Figure 5.9 R20 The Fresnel equations are
rs≡ Er(s)
Ei(s) = sin θtcos θi− sin θicos θt
sin θtcos θi+ sin θicos θt
ts≡ Et(s)
Ei(s) = 2 sin θtcos θi sin θtcos θi+ sin θicos θt rp ≡ Er(p)
Ei(p) = cos θtsin θt− cos θisin θi
cos θtsin θt+ cos θisin θi
127
tp ≡ Et(p)
Ei(p) = 2 cos θisin θt
cos θtsin θt+ cos θisin θi
(a) Find what each of these equations reduces to when θi= 0. Give your answer in terms of ni and nt.
(b) What percent of light (intensity) reflects from a glass surface (n = 1.5) when light enters from air (n = 1) at normal incidence?
(c) What percent of light reflects from a glass surface when light exits into air at normal incidence?
R21 Light goes through a glass prism with optical index n = 1.55. The light enters at Brewster’s angle and exits at normal incidence.
Figure 5.10
(a) Derive and calculate Brewster’s angle θB. You may use the results of R19 (c).
(b) Calculate φ.
(c) What percent of the light (power) goes all the way through the prism if it is p-polarized? Ignore light that might make multiple reflections within the prism and come out with directions other than that shown by the arrow. You may use the Fresnel coefficients given in R20.
(d) What percent for s-polarized light?
R22 A 45◦- 90◦- 45◦prism is a good device for reflecting a beam of light parallel to the initial beam. The exiting beam will be parallel to the entering beam even when the incoming beam is not normal to the front surface (although it needs to be in the plane of the drawing).
(a) How large an angle θ can be tolerated before there is no longer total internal reflection at both interior surfaces? Assume n = 1 outside of the prism and n = 1.5 inside.
Figure 5.11
(b) If the light enters and leaves the prism at normal incidence, what will the difference in phase be between the s and p-polarizations? You may use the Fresnel coefficients given in R20.
R23 Second harmonic generation (the conversion of light with frequency ω into light with frequency 2ω) can occur when very intense laser light travels in a material.
For good harmonic production, the laser light and the second harmonic light need to travel at the same speed in the material. In other words, both frequencies need to have the same index of refraction so that harmonic light produced down stream joins in phase with the harmonic light produced up stream, referred to as phase matching. This ensures a coherent building of the second harmonic field rather than destructive cancellations.
Unfortunately, the index of refraction is almost never the same for different fre-quencies in a given material, owing to dispersion. However, we can achieve phase matching in some crystals where one frequency propagates as an ordinary wave and the other propagates as an extraordinary wave. We cause the two indices to be precisely the same by tuning the angle of the crystal.
Consider a ruby laser propagating and generating the second harmonic in a uni-axial KDP crystal (potassium dihydrogen phosphate). The indices of refraction are given by no and
none
pn2osin2φ + n2ecos2φ
where φ is the angle made with the optic axis. At the frequency of a ruby laser, KDP has indices no(ω) = 1.505 and ne(ω) = 1.465. At the frequency of the second harmonic, the indices are no(2ω) = 1.534 and ne(2ω) = 1.487.
Show that phase matching can be achieved if the laser is polarized so that it experiences only the ordinary index and the second harmonic light is polarized perpendicular to that. At what angle φ does this phase matching occur?
Selected Answers
R17: (b) 1/4, (c) 1/2.
129 R20: (b) 4% (c) 4%.
R21: (b) 33◦, (c) 95%, (d) 79%.
R22: (a) 4.8◦, (b) 74◦. R23: 51.12◦.