SIMG Solution Set #5

Transcription

1 SIMG Solution Set #5. Describe completely te state of polarization of eac of te following waves: (a) E [z,t] =ˆxE 0 cos [k 0 z ω 0 t] ŷe 0 cos [k 0 z ω 0 t] Bot components are traveling down te z-axis towards +. Te two polarizations ave te same magnitude and te same pase, but note tat te y-component is multiplied by. Tis electric field could be rewritten as: E [z, t] = ˆxE 0 cos [k 0 z ω 0 t] ŷe 0 cos [k 0 z ω 0 t] = ˆxE 0 cos [k 0 z ω 0 t]+ŷe 0 cos [k 0 z ω 0 t π] Tus wen te x-component reaces its maximum positive value, te y-component reaces its negative extremum. Te state of polarization is linear and oriented at an angle of 5 measured from te x-axis. (b) E [z,t] =ˆxE 0 sin π z λ 0 ν 0 t ŷe 0 sin π z λ 0 ν 0 t Tis is just a disguised version of te previous question. Apply te definitions for te wavevector k and te angular temporal frequency ω : E [z, t] = ˆxE 0 sin π z λ 0 ν 0 t ŷe 0 sin π z λ 0 ν 0 t = ˆxE 0 sin [k 0 z ω 0 t] ŷe 0 sin [k 0 z ω 0 t] = ˆxE 0 cos k 0 z ω 0 t π i ŷe 0 cos k 0 z ω 0 t π i = ˆxE 0 cos k 0 z ω 0 t π i + ŷe 0 cos k 0 z ω 0 t π i + π = ˆxE 0 cos k 0 z ω 0 t π i + ŷe 0 cos k 0 z ω 0 t + π i Te state of polarization again is linear and oriented θ = 5 measured from te x-axis. (c) E [z,t] =ˆxE 0 sin [ω 0 t k 0 z]+ŷe 0 sin ω 0 t k 0 z π Note tat te component waves now travel towards. I suggest rewriting in terms of cosine functions, toug tis is not necessary: E [z,t] = ˆxE 0 sin [ω 0 t k 0 z]+ŷe 0 sin ω 0 t k 0 z π i = ˆxE 0 cos ω 0 t k 0 z π i + ŷe 0 cos ω 0 t k 0 z π π i = ˆxE 0 cos ω 0 t k 0 z π i + ŷe 0 cos ω 0 t k 0 z 3π ³ = ˆxE 0 cos ω 0 t k 0 z π i µ + ŷe 0 cos ω 0 t k 0 z 3π = ˆxE 0 cos k 0 z ω 0 t + π i + ŷe 0 cos k 0 z ω 0 t + 3π so te pase of te y-component is smaller tan tat of te x-component by π radians. Te two magnitudes are identically E 0, but te pase cange is not a

2 multiple of π radians (necessary for linear polarization) or π radians (for circular polarization), so te state of polarization must be elliptical. We need to determine te andedness. If we use te angular momentum convention (my preference), ten te field observed at a fixed z looking back at te source rotates counterclockwise if te polarization is rigt-anded. Note tat in te so-called optics or screwy convention, ten a field tat is rigt-and elliptically polarized (RHEP) field rotates rclockwise. Look at te field for a fixed value of z (say z =0): E [z,t] = ˆxE 0 cos ω 0 t + π i + ŷe 0 cos ω 0 t + 3π = ˆxE 0 cos ω 0 t π i + ŷe 0 cos ω 0 t 3π Te Argand diagram of tis field sows tat it rotates counterclockwise, so te field is RHEP in te angular momentum convention. (d) E [z,t] =ˆxE 0 sin [ω 0 t k 0 z]+ŷe 0 sin ω 0 t k 0 z + π Again, tis wave travels towards. Rewrite as cosines: E [z, t] = ˆxE 0 sin [ω 0 t k 0 z]+ŷe 0 sin ω 0 t k 0 z + π i = ˆxE 0 cos ω 0 t k 0 z π i ³ + ŷe 0 cos ω 0 t k 0 z + π π i = ˆxE 0 cos k 0 z ω 0 t + π i + ŷe 0 cos [k 0 z ω 0 t] Again, te magnitudes of te two components are equal, but te pase difference π is an odd multiple of radians, so te polarization is circular. For andedness, look back at te source from a fixed location (z=0) and determine wic direction te field rotates: E [z,t] = ˆxE 0 cos ω 0 t + π i + ŷe 0 cos [ ω 0 t] = ˆxE 0 cos ω 0 t π i + ŷe 0 cos [ω 0 t] Again, te E vector rotates counterclockwise, so te polarization is RHCP in te angular-momentum convention.

3 . A beam of linearly polarized ligt is incident on an ideal linear polarizer. Find te angle between te axis of te directlion of oscillation of te electric field and te ideal linear polarizer if te 5% of te irradiance (or intensity ) is transmitted. Te polarizer passes te portion of te electric field amplitude projected onto te axis of te polarizer (see figure), so tat te field becomes: E θ = E 0 cos [θ] Te irradiance of te ligt is te time average of te squared magnitude of te electric field: I θ = E θ = E 0 cos [θ] =I 0 cos [θ] So if I θ = I 0, ten: # θ =cos "r = π 3 radians =60 3

4 3. Sketc a diagram of te windsield and dasboard of a car and use it to explain te useful direction of polarization of polaroid sunglasses. So te ligt tat is reflected from te dasboard tat is reflectedintoteeyeispreferentially polarized perpendicular to te plane of te page, i.e., orizontally as seen by te eye. We would like to ave te sunglasses block tis polarization and pass te ligt tat is polarized in te plane of te page, i.e., vertically as seen by te eye.

5 . Prove Malus law tat te intensity transmitted by a pair of ideal linear polarizers oriented wit teir axes at angle θ 0 is proportional to cos [θ 0 ]. Tis is just a variation of #. If unpolarized ligt wit electric field E 0 passes troug one polarizer at an arbitrary angle, ten te emerging electric field from te polarizer is E = Z + π π E 0 cos [θ] dθ = E 0 and te emerging irradiance is: I E Te electric field is incident on a second polarizer oriented at angle θ measured relative to te first polarizer, so te emerging electric field is: E = E cos [θ] = E 0 Te emerging irradiance is te square: cos [θ] I E = E cos [θ] = E 0 cos [θ] cos [θ] 5

6 5. Consider two ideal linear polarizers tat are placed wit teir polarizations axes ortogonal. A tird ideal linear polarizer is placed betweent tese two and rotated at a constant rate ω 0 [radians per second]. Sow tat te emerging ligt intensity oscillates at four times te rotation frequency. If te middle polarizer is oriented along te axis of eiter of te oter polarizers, ten te output is zero because te ligt troug te parallel pair of polarizers (say tat tree times fast) is blocked by te ortogonal polarizer. So some ligt is passed only wen te middle polarizer is oriented at an angle θ tat differs from te oter two. Consider te first two polarizers te output is just tat from Malus law: E = E cos [θ] Te angle between te axes of te second and tird polarizer is φ = π θ, so te amplitude emerging from te tird polarizer may be easily expressed in terms of te amplitude out of te second: π i E 3 = E cos [φ] =E cos θ ³ π i π i = E cos cos [θ]+sin sin [θ] = E sin [θ] = (E cos [θ]) sin [θ] BUT, θ = ω 0 t: = E BUT, sin [α] = ( cos [α]): sin [θ] cos [θ] = E sin [θ] E 3 = E sin [ω 0t] I 3 E 3 = E sin [ω 0 t] I 3 E cos [ ω 0 t] = E cos [(ω 0 ) t] 8 wic oscillates at an angular temporal frequency of ω 0 radians sec. 6

7 6. Te wavelengt of unpolarized ligt from a mercury source is λ =56.07 nm. If incident on a plate of glass at an angle θ i =58 0 0,tereflected ligt is seen to be completely linear polarized. Find te refractive index of te glass. If ligt is reflected obliquely from te sufrace of a transparent material, ten bot te reflected and transparent beams are (in general) partially polarized. If incident at Brewster s angle, te reflected beam is completely plane polarized. Tis angle is specified by Brewster s law, wic evalutes tis wavelengt to: n = tan[θ B ] n n = tan[ ] ' A source of left-and circularly polarized ligt at λ 0 =656nmsould be converted to rigt-and circularly polarized ligt by passing it troug a tickness of quartz (SiO ), wic as n s =.55 and n f =.5. (a) Compute te minimum tickness of a plate tat will accomplis te task. We want to create a pase delay of cycle, or π radians between te two polarizations via te difference in optical pat lengts. For a plate wit pysical tickness d, te optical pat lengt is nd, so te optical pat lengts in te two directions are n s dandn f d. Te difference in optical pat lengt is Te difference in te index of refraction in te two directions is: (n s n f ) d =(.55.5) d =0.009d Te difference in te number of wavelengts is te optical pat difference divided by te vacuum wavelengt λ 0 : (n s n f ) d λ 0 =0.009 d λ 0 = difference in te number of wavelengts To get a pase delay of π radians, tis difference as to be π = wavelengt: π = d = d = µ 656 nm =36, nm = 55 5 λ λ 0 ' mm wic is pretty tin (b) (EXTRA CREDIT) From te result of (a), you can see tat suc a plate is not very practical. Modify te design of te plate to create a practical device. Tis plate is very tin. We can create a practical pase plate by constructing a sandwic of two suc plates oriented at 90 to eac oter so tat one generates a pase difference of πn +π radians in one direction, were N is a large number, wile te oter generates πn radians of pase difference in te oter direction. Te sandwic can be quite tick, but te resulting pase difference is π radians. 7