Edward S. Rogers Sr. Department of Electrical and Computer Engineering ECE318S Fundamentals of Optics Midterm II Mar. 21, 217 18:3 2: Exam Type: C (A doublesided aid sheet and a nonprogrammable calculator are allowed) Exam Duration: 9 minutes Q1 Q2 Q3 Total 2 2 2 6 Note: (1) Your numerical answers should be followed by appropriate units, whenever applicable. (2) Follow the sign conventions established in the lectures. (3) Unless explicitly stated in the question, explanations to your answers are optional. However, partial marks will be awarded for incorrect or incomplete answers when accompanied by correct and relevant explanations. Conversely, marks will be deducted accordingly if a wrong explanation accompanies a correct answer. (4) If, in your opinion, a question is unclear or information is missing, then make reasonable assumptions, state them in your answer, and proceed.
ECE318S Fundamentals of Optics Midterm II Page 2 of 8 Q1 You have only two quarterwave plates and you wish to change the polarization of a collimated, horizontally polarized laser beam. Take the horizontal direction as x, and the beam propagation direction as z. Specify the number (one or two) of waveplates you will need, and the direction of the slow axis of each of the waveplates you will use in order to (1) [3 marks] Convert the laser to a circular polarization (2) [5 marks] Convert the laser to a vertical (linear) polarization (3) [5 marks] Convert the laser to a linear polarization in the cosα x + sinα y direction, where <α<π. (4) [7 marks] Write down the Jones Matrix of your polarization conversion system used in (3), in the given coordinate system. Solution: (1) Need one quarterwave plate centered on and normal to the beam path, and its slow axis should be ±45 o or ±135 o with respect to the x axis, on the xy plane. (2) Need two quarterwave plates centered on and normal to the beam path. The slow axes of both waveplates should be aligned with each other, making the combination effectively a halfwave plate, and the slow axis should be ±45 o or ±135 o with respect to the x axis, on the xy plane. (3) Need two quarterwave plates centered on and normal to the beam path. The slow axes of both waveplates should be aligned with each other, making the combination effectively a halfwave plate, and the slow axis should be ±α/2 or ± (α+π)/2 (3 marks) with respect to the x axis, on the xy plane. (4) Method 1: The Jones Matrix for the combination of two quarterwave plates with both slow axes aligned to α/2 angle with respect to the xaxis, is given by: M = R(!!)T QWP T QWP R(!) ( cos!! 2 sin!! + = 2 (!sin!! 2 cos!! 2 ) ), ( = )!cos!sin!sin cos +, e i! /2 1 + (, ) e i! /2 1 ( + cos 2,!sin ) 2 sin 2 cos 2 Note, given the other possibilities, any of the following Matrices are also OK:!cos! sin! sin! cos! cos! sin! sin!!cos! cos!!sin!!sin!!cos! +,
ECE318S Fundamentals of Optics Midterm II Page 3 of 8 Method 2: If the input Jones Vector is [1 ] T and the output Jones Vector is [cosα, sinα] T, then the matrix M relating them must satisfy:! cos!! = M 1 sin! Also, M must be unitary (lossless system). Therefore, M = cos! sin! sin!!cos!
ECE318S Fundamentals of Optics Midterm II Page 4 of 8 Q2 A collimated, vertically (y direction) polarized, laser beam is passing through a Faraday rotator followed by a linear polarizer with a vertical (y) transmission axis. The Faraday rotator is 2cm long, has a Verdet constant of 134 rad/(tm) at the laser wavelength, and is subject to a uniform, timevarying magnetic field expressed as B =.5sin(1πt) z Tesla, where t is time in seconds and z is the beam propagating direction. (1) [3 marks] What is the Jones vector representation of the polarization before entering the rotator? (2) [5 marks] Plot the polarization rotation angle experienced by the light beam after passing through the rotator as a function of time. Label the horizontal scale on the time axis and the maximum polarization rotation angle on the vertical scale. (3) [6 marks] What is the Jones vector representation of the polarization after exiting the rotator? (Hint: it should be a function of time.) (4) [6 marks] Assuming the input intensity of the beam is I o and the rotator is lossless, what is the intensity of the laser beam after passing through the polarizer? Express it as a function of time. Solution: (1) The Jones Vector for vertical polarization is [ 1] T. (2) The rotation angle imposed by the Faraday rotator when B is parallel to the direction of propagation is θ(t)=vbd=( 134)x[.5sin(1πt)]x.2= 1.34sin(1πt) rad. (3) Jones Matrix for the Faraday Rotator is
ECE318S Fundamentals of Optics Midterm II Page 5 of 8 M = = = cosvbd!sinvbd sinvbd cosvbd cos[!1.34sin(1!t)]!sin[!1.34sin(1!t)] sin[!1.34sin(1!t)] cos[1.34sin(1!t)]!sin[1.34sin(1!t)] cos[!1.34sin(1!t)] sin[1.34sin(1!t)] cos[1.34sin(1!t)] Therefore the Jones vector after exiting the rotator is: cos[1.34sin(1!t)] J rot _ out = MJ in =!sin[1.34sin(1!t)] sin[1.34sin(1!t)] =!cos[1.34sin(1!t)] sin[1.34sin(1!t)] cos[1.34sin(1!t)] 1 (4) The Jones vector after passing through the polarizer is :! J out = M pol _ y J rot _ out =! sin[1.34sin(1!t)] 1 cos[1.34sin(1!t)]! = cos[1.34sin(1!t)] Therefore the intensity after passing through the polarizer is: I out = I o cos 2 [1.34sin(12!t)]
ECE318S Fundamentals of Optics Midterm II Page 6 of 8 Q3. Three identical right angled glass prisms are used as reflectors in a periscope set up shown below. After three 9degree turns, the beam emerges in the same propagation direction as the input propagation direction. The beam enters and exits all prisms at normal incidence to the facets, and reflects internally at 45 o incidence in each prism. The refractive index of the prism is n g =1.5. The medium outside the prisms is air, and the glass material is lossless. The input is linearly polarized in the x direction. (1) [5 marks] Determine the overall transmittance of the system consisting of the three prisms. Transmittance is defined as the ratio of output power to input power. (2) [5 marks] Determine the output polarization state and express it by a Jones vector in the xyz coordinate system shown. (3) [5 marks] Determine the Jones Matrix representation of the system. (4) [5 marks] Repeat (1) for the case where the medium outside the prisms is water with refractive index n w =1.33. Solution: (1) For each prism, the beam goes through 2 normalincidence transmissions at the facets, and one total internal reflection (TIR) at 45degree incidence. (The critical angle is θ c =sin 1 (1/n g )=41.8 o ). For each normalincidence transmission, the transmittance is T facet =1[(n g 1)/(n g +1)] 2 =.96. For each total internal reflection, the transmittance is 1. Therefore, the total transmittance of the system is T 6 =.78. (2) Three qualitative remarks here: (a) The six normalincidence transmissions do not change the polarization state. So we only need to take into account the three TIRs. (b) The plane of incidence for the first TIR (xz plane) is orthogonal to the plane of incidence for the second TIR (xy plane), which is orthogonal to the plane of incidence for the third TIR (yz plane). Therefore there is a role change between TE and TM polarization in between the prisms. For example, the TM output polarization of the first TIR becomes the TE input polarization of the second TIR, and the TE output
ECE318S Fundamentals of Optics Midterm II Page 7 of 8 polarization of the second TIR becomes the TM input polarization of the third TIR. (c) According to our sign convention, all blue E fields indicated in the diagram below are related to each other by the Fresnel coefficients, so are all the red E fields. Notice here, if the input is polarized in the xdirection, the output will be polarized in the y direction. Therefore the output polarization state is: J out =!1 or J out = 1 This is a linearly polarized state in the y direction. (3) Consider the diagram above, we note that the xcomponent of the output E field depends only on the ycomponent of the input E field, and vice versa. Also, the transmissions at normal incidence do not alter the polarization state, so we need only consider the three TIRs. Therefore, T =!r TE r TM r TE!r TM r TE r TM =!exp i! +! +! TE TM TE!exp i (! TM +! TE +! TM ) ( ) where n! TE =!2 tan!1 g sin 45 o n g cos45 o ( ) 2!1 =!.6435rad =!36.9 o
ECE318S Fundamentals of Optics Midterm II Page 8 of 8 n! TM =!2tan!1 g n g sin 45 o cos45 o ( ) 2!1 +! =1.85rad =16.3 o Therefore, T = [ ]!exp i.56!exp[ i3.6] or 1 exp( i2.5) or exp (!i2.5) 1 (4) If the surrounding medium is water, TIR will not occur because the critical angle for the waterglass interface is θ c = sin 1 (n w /n g ) = 62.5 o. With θ i =45 o, θ t = sin 1 (n i sinθ i /n t ) = 52.9 o. The Fresnel reflection coefficients for TE and TM waves are: r TE = n g cos45o! n w cos52.9 o n g cos45 o + n w cos52.9 o =.14 r TM =!n w cos45o + n g cos52.9 o n w cos45 o + n g cos52.9 o =!.19 For each normal incidence, the transmittance is T facet =1[(n g n w )/(n g + n w )] 2 =.996. For each 45degree reflection, the TE reflectance is R TE =r TE 2 =(.14) 2 =.192, and the TM reflectance is R TM =r TM 2 =(.19) 2 =3.7x1 4. Therefore, the overall transmittance for an xpolarized input is: T overall = T 6 R TM R TE R TM =.996 6 3.7!1 4 ( ) 2.192 = 2.6!1 9 In other words, essentially no transmission when the surrounding is water! (If you are going to use this periscope underwater, better put it in a sealed, waterproof enclosure.)