Solution Set 1 Phys 4510 Optics Fall 2013

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Damian Craig
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1 Solution Set Phys 450 Optics Fall 203 Due date: Tu, September 7, in class Reading: Fowles , Derive the formulas: and (see Fowles problem.6) v g = v φ λ dv φ dλ = λ 0 dn (2) v g v φ c dλ 0 () v g = dω dk = d(v φk) = v φ + k dv φ dk dk = v φ + k dv φ dλ dλ dk Using the relation between λ = 2π dλ k we get for dk = 2π k and thus for v 2 φ v g = v φ + k 2π k 2 dv φ dλ = v φ 2π k dv φ dλ = v φ λ dv φ dλ So we have proven the first relation. The second is just as straightforward. = dk v g dω = d ( nω ) = n dω c c + ω dn c dω = + ω dn dλ 0 v φ c dλ 0 dω Now we have to be careful not to mix up the wavelength in vacuum λ 0 and wavelength in matter λ = λ0 n. Using the relation between the wavelength and the wavenumber λ 0 = 2π k o = 2πc ω we get dλ 0 ω = 2πc ω. This yields 2 v g = v φ + ω c dn dλ 0 ( 2πc ) dω 2 = λ 0 v φ c dn dλ 0 2. The Cauchy equation provides a good first order approximation of the wavelength dependence of the refractive index of transparent media given by: n(λ 0 ) = B + C λ 0 2 (3) with the empirical coefficients B and C. For borosilicate glass (BK7, a glass often used for lenses and other optical components for the visible spectral range), B =.5046 and C = µm 2. Calculate phase and group velocity for λ 0 = 400, 532, 632, and 800 nm. Discuss the trend. (after Fowles problem.7) Using the second equation from the previous task we can make the following table: λ 0 [nm] n(λ 0 ) v φ v g c 0.632c c 0.646c c 0.65c c 0.656c

2 From the equation for the refractive index we can see that for increasing λ the refractive index tends towards B. Thus, the group velocity tends towards the phase velocity for increasing wavelength. It will, however, never exceed the phase velocity. 3. Under which circumstances can the phase velocity in a medium exceed the group velocity? Why is this not in contradiction to special theory of relativity? From equation 2 we can see that for the phase velocity to be greater than the group velocity the derivative of the refractive index with respect to the wavelength has to be positive. This is known as anomalous dispersion. Since the phase velocity cannot transport information it is allowed to travel faster than the group velocity or even the speed of light since only the transport of information fast than the speed of light is prohibitted by special theory of relativity. 4. A scuba diver, at a depth of d = 0 m below the water surface, when looking up sees a circular region through which he can look outside the water. The index of refraction of water is n =.33. Provide corresponding sketches with your solutions to the following problems: n ϕ ϕ S ϕ 2 ϕ DS n 2 Figure : Schematic drawing of the diver looking at the sunset above the water and looking at his/her circular area of view above the water. a) Suppose the sun is at φ S = 45 above the horizon as seen by an observer outside the water. Under what angle with respect to the water surface would the diver see the sun from his vantage point 0 m below the surface? Using Snell s law n sin(φ ) = n 2 sin(φ 2 ) for φ = 45 = 90 φ S and n =, n 2 =.33, we get φ 2 = 32.. The angle under which the diver observes the sunset however is the angle φ DS with the horizontal. Thus, we get φ DS = b) What would be the angle φ max under which the diver would observe sun set? 2

3 Using Snell s law for φ = 90 φ S where φ S = 0 for the sunset now and n =, n 2 =.33, we get φ 2 = The angle under which the diver observes the sunset however is the angle φ DS with the horizontal. Thus, we get φ DS = 4.2. c) How large is the radius r of that circular field of view of the diver? What does the driver see when looking in other directions? For this task the angle of total internal reflection is of itnerest. It can be calculated by calculing for what angle of φ 2 sin(φ ) becomes. This is the case for φ = 90. Thus the angle of total internal reflection is as calculated in b) φ 2 = 48.8 circ. Thus, get for the radius of the divers field of view.4m. If the diver looks towards the surface at a bigger angle than the angle of total internal reflection he will see reflections of the life under water. 5. We want to study reflection and refraction of unpolarized light impinging onto a glass surface in air. Consider two different colors: red light at λ r = 656 nm subject to a index of refraction of n r =.5076, and violet light at λ v = 405 nm with n v = a) How large is the reflectivity of the surface for normal incidence for both wavelengths? For normal incidence of the incoming light we know that the formular for the reflectivity reduces to ( n2 ) 2 n R = n 2 n + So we get for R r = 4.0% and for R v = 4.30%. b) Under what angle of incidence is the reflected light completely polarized? The angle under which the incoming light is compeltely polarized is the Brewster angle. θ Brewster = arctan( n 2 n ) so that we get θ Brewsterr = 56.4 and θ Brewsterv = 56.7 c) How large is the critical angle α c for total internal reflection? The critical angle above which total internal reflection in the medium occurs is the angle for which the following is true ( ) n α c = arcsin Thus, we get α r = 4.6 and α v = Assume we have a perfect polarizer by which we mean a piece of material which passes all light polarized along the axis of the polarizer and none of the light perpendicular to this axis. n 2 3

4 a) Assume we initially have unpolarized light, how much light (i.e. in terms of intensity) is passed if we have two succesive polarizers with an angle α between the two? Unpolarized light is characterized by equal amounts of two orthogonally polarized components. In other words, the intensity averaged over time is equal if we pass the light through a polarizer oriented along the x-axis or the y-axis. So the first polarizer passes half the incident intensity I 0. Now, a polarizer has the effect of projecting out and passing the incident field along the polarizer axis while absorbing or reflecting the other component. Thus the electric field passed will be the incident field modulated by the cosine of the angle between the incident field and the polarizer axis. Hence after passing through the second polarizer, the intensity is I(α) = I 0 2 cos2 (α). Note that the above equation in a slightly different form is referred to as Malus law. b) Now, what happens if we place a third polarizer after the second with its axis perpendicular to the first? Plot the transmitted intensity as a function of the angle of the second polarizer. Can you think of an analogous situtation in quantum mechanics? Working along the same lines as in part (a) we find I(α) = I 0 2 cos2 (α) sin 2 (α). A plot of the transmitted intensity is shown below demonstrating that unless α = 0 or α = π/2 there is nonzero transmission. This problem is very analogous to the Stern-Gerlach experiment in which the apparatus acts essentially as an electron spin polarizer I I Figure 2: Plot of transmitted intensity versus polarizer angle for a three polarizer system with the first and last cross polarized and the middle at angle α with respect to the first. Α 4

5 c) Finally, suppose we have N polarizers each rotated from the last by an angle φ = φ/n. How much intensity is passed in the limit that N? What does this system do? Hint: You may need L Hopital s rule... We apply Malus law N times with the angle between successive polarizers φ = φ/n giving I N (φ) = I 0 2 cosn φ. In the limit that N we find the indeterminant form, but after applying a logarithm and L Hopitals rule we find the limit of the cosine term equals so that I (φ) = I 0 2. In other words, the system rotates the polarization without extinguishing any of the light polarized along the axis of the first polarizer. 7. NASA proposes to launch a solar sail in 204. a) Describe in words along with the appropriate equations the physical principle behind this idea. Making appropriate simplifying assumptions (e.g. ignoring planetary perturbations and assuming radial motion), consider the design parameters of such a sail. For example, what is the maximum mass per unit area of the sail? How do your design parameters compare to those given in the link from the footnote below? Do the discrepancies make sense? Light carries momentum, and when it scatters off of a reflecting surface, its mometum is changed. The amount by which it is changed must be balanced by the momentum gained by the reflecting surface. This means that light exerts a force, which, upon dividing out the area of the surface from which the light is reflected gives a radiation pressure for a flat surface of P = 2I c cos2 (θ), where θ is the angle the light makes with the surface normal. This equation can be derived from knowing the number of photons hitting the surface each second, the average energy of the photons, the energy-momentum relation for light, and realizing that only the component of momentum perpendicular to the surface changes sign (factor of 2 in numerator). Now, suppose we want to operate at threshold, this means that gravity exactly balances the radiation force so that Energy conservation gives 2I(r)A sail c = GM m sail r 2. I(r) = I r 2 r 2, where I 366W/m 2 and r are the intensity of sunlight at Earth and the radius of Earth s orbit around the sun respectively. Thus the maximum areal mass density denoted σ is σ = m sail A sail = 2I r 2 cgm. Numerically this turns out to about σ =.6g/m 2. NASA s sail at 70lbs 32kg and with 3000ft 2 200m 2 gives σ NASA 20g/m 2. So, as it turns out, NASA can get away with a See and included links 5

6 factor of 20 larger areal mass density. Why is that? Well, it appears that NASA intends to use their craft for a space weather warning system rather than interstellar travel, which must be the reason they can get away with this. b) Suppose we want to send this craft to the orbit of Pluto (sorry, no longer a planet...) from Earth. Again ignoring planetary perturbations and assuming radial motion, how long would this take using the NASA spacecraft parameters? Hint: I strongly recommend using energy conservation, and be careful with the potential energy! Well, it turns out that the NASA parameters will not allow for us to ever get to Pluto without an external illumination source (e.g. a laser) but it seems that the required power is enourmous, so those dreams are probably a ways off. So if you stopped here, you have technically answered the question correctly. But if you are curious and want to give it a go with a magic material with a factor of say 0 better mass density than threshold, then we can get somewhere. There is an effective potential energy of U(r) = 2I r 2 A sail cr GM m sail, r so that assuming zero initial velocity and applying energy conservation gives t = rp r 2dr m U(r ) U(r) Calculating this gives about t =.6yr. That seems insanely fast, infact the average velocity is about 5km/s according to this calculation. However, say we made our sail from something like aluminum which has ρ = 2.70g/cm 3 near room temperature. This sheet would have to be less than 60nm thick to get σ 0 /0! The skin depth for aluminum is around 3 5nm in the visible spectrum, so we could expect that the sail would still be fairly reflective, however, manufactruing such a sheet seems rather unfeasible. So unless we have an additional source of light to propel the sail, it seems this idea is not likely to lead to spacecraft designed for exploring beyond Earth s orbit. 6

Solution Set 2 Phys 4510 Optics Fall 2014

Solution Set 2 Phys 4510 Optics Fall 2014 Solution Set Phys 4510 Optics Fall 014 Due date: Tu, September 16, in class Scoring rubric 4 points/sub-problem, total: 40 points 3: Small mistake in calculation or formula : Correct formula but calculation