List of In-Class Assignments Solutions

Sept. 11, 2017
 List 2 properties of laser light. Coherent (temporal and spatial), monochromatic, stable phase, ideally TEM_{0 0} and directional without spreading. Extra-credit A laser lases at 3 mu m with a gain profile having linewidth of 10 nm. Find bandwidth of the gain profile nu = c_o / lambda_o = {3 times 10^8} / {3 times 10^-6} = 10^{14} Hz {Delta nu} / nu = {Delta lambda} / lambda => Delta nu = nu {Delta lambda} / lambda =>10^14 {10 nm}/ {3000 nm} = 1/3 times 10^12 (Hz) = 1/3 (THz)

Sept. 13, 2017
 a) List the 3 major components of a laser. Pumping (energy) source, gain (active) medium, feedback. b) Gain profile has bandwidth of 5 THz and the free spectral range Delta nu_{FSR} = 20 GHz. Estimate the number of longitudinal modes in the laser? Number of longitudinal modes = {Delta nu_g} / {Delta nu_{FSR}}=5000/20=250. c) A pulse laser lases at 3 mu m with a gain profile having linewidth of 10 nm. It emits square pulse with energy of 1 mu J per pulse, pulse width of 1 fs and repetition rate of 500 Hz. i) Find photon energy of the laser in eV. E(ev) = 1.24 / 3 = 0.413 (eV). ii) Find the peak power of the laser. P_{peak} = E_{"pulse"} / t_{"pulse"} = 10^{-6} / 10^{-15} = 10^9 (W) = 1 (GW). Extra-credit iii) Find average power of the laser in c). P_{peak} = E_{"pulse"} times repetition rate = 10^{-6} times 500 = 5 times 10^{-4} (W) = 0.5 (mW). iv) Estimate the number of longitudinal modes for the laser in c) if Delta nu_{FSR} = 10 GHz. From the extra-credit of last in-class (Sept 11), Delta nu_{g} = 1/3 (THz). Number of longitudinal modes = {1/3 times 10^{12}} / 10^(10} = 100/3 = 33 modes. Notice that this number must be in integer.

Sept. 20, 2017
 vec ccH = hat z cos (10^6 t + 1/300 y ) a) Identify propagation direction. b) Find speed of the wave. c) Find polarization direction of the wave. Extra-credit If the medium where the wave propagate has mu = mu_o= 4 pi times 10^{-7} H/m, find the permittivity epsilon of the medium. a) The propagation direction is the spatial variable(s) inside the cos or sin functions. In this case, it is: hat k = - hat y. In general, the argument inside the cos or sin function is omega t - vec k cdot vec r = omega t - k hat {k} cdot vec r. For this problem hat k cdot vec r = - hat y cdot (x hat x + y hat y + z hat z)=-y. However, in general hat k cdot vec r = {vec k}/ | vec k| cdot ( x hat x + y hat y + z hat z ) = 1/k (k_x hat x + k_y hat y + k_z hat z) cdot ( x hat {x} + y hat {y} + z hat {z} ) = {k_x x + k_y y + k_z z}/k where k = | vec k | = sqrt {k_x^2+k_y^2+k_z^2}. For example, cos ( t + y + x) states that omega = 1, k_y = 1 and k_x =1 . Hence, vec k = - hat x - hat y and hat k = -1/sqrt{2} hat x - 1/sqrt{2} hat y, i.e. propagating in a direction making angle of 225^o  from the positive x axis. b) c = omega / k = {10^6} / (1/300} = 3 times 10^8 m/s = c_o c) Polarization direction is hat E = hat {H} times hat k = hat z times - hat y = hat x If the propagation direction is along one of the coordinate axes, polarization direction should be the axis not along the H direction and the propagation direction, i.e. the axis not mentioned in the H expression. In general, we should use hat {H} times hat k to find hat {E}, particularly for propagation direction between axes or off-axis. Extra-credit Since c = c_o and mu = mu_o, epsilon = epsilon_o. c = 1 / sqrt{ mu epsilon} => epsilon = 1 / {mu c^2} = 1/ {36 pi times 10^{9}} = 8.84 times 10^{-12} F/m

Sept. 27, 2017
Sept 27, 2017 solution

Oct. 4, 2017
 1) Consider n_1=1, n_2=1.5. Find the power reflectivity at boundaries 1 and 2. Boundary 1: R = | (n_1-n_2)/(n_1+n_2) |^2 = |0.5/(2.5)|^2 = 1/25 Boundary 2: R = | (n_2-n_1)/(n_1+n_2) |^2 = |0.5/(2.5)|^2 = 1/25. 2) What are the two special ray conditions for finding the ray matrix. Condition 1: r=0; condition 2: r prime =0. 3) How many ray matrices are required for relating Ray in and Ray out. There are 2 boundaries and 1 space => 3 ray matrices. Extra-credit 4) Write down all matrices required in 3). [(1,0),(0,n_2/n_1)] [(1,L), (0,1)] [(1,0),(0,n_1/n_2)] =[(1,0),(0,1.5)] [(1,L), (0,1)] [(1,0),(0,1/1.5)]

Oct. 9, 2017
 Find the ray matrix for the system on right. [(1, d_1),(-1/{f_1} , 1-d_1/{f_1})] [(1, d_2),(-1/{f_2} , 1-d_2/{f_2})] =[(1 - d_1/{f_2} , d_2+ d_1 (1 - d_2/{f_2})),(-1/{f_1}-1/{f_2} (1-d_1/{f_1}) , -d_2/{f_1}+(1-d_1/{f_1})(1-d_2/{f_2}))] Ectra-credit Draw the equivalent diagram in terms of lenses for the mirrors on the right and indicates the unit cell.

Oct. 11, 2017
 a) Draw the equivalent lens diagram for the cavity on the right. b) Identify unit cell at the sample plane Extra-credit Is the resonator stable? [(A,B),(C,D)]=[(1,1),(0,1)][(1,1),(-1/2,1 - 1/2)]=[(1/2,3/2),(-1/2,1/2)] {A+D}/2 = 1/2 (1/2 + 1/2) = 1/2 < 1 => The resonator is stable.

Oct. 16, 2017
 1) Find Delta nu_{FSR} and power reflectivity at the end of the etalon (FP cavity) on the right. Delta nu_{FSR} or nu_F = c_o / {2nd} = {3 times 10^{8}}/{2 times 5 times0.01} = 3 times 10^9 (Hz). Power reflectivity frR_1 = frR_2 = frR = |n_1 - n_2|^2/|n_1 + n_2|^2 = 4^2/6^2 = 4/9. Extra-credit For 1), Finesse = ? frF = {pi (frR frR)^(1/4))/(1-sqrt(frR frR)} = {pi (frR )^(1/2))/(1-frR }=pi( 2/3)/ (5/9)={6pi}/5.
Oct. 18, 2017
 1) Find the fractional power transmitted from one end to the other end. T = (1- frR)^2 = (1- 4/9)^2 = 25/(81)=0.309 Extra-credit If the block of material has attenuation coefficient alpha = 50 (m^(-1)), now what is the fractional power transmitted? T = (1- frR)^2 e^(-50 times 0.01) = 25/(81) e^{-0.5}=0.1872
Oct. 23, 2017
 1) a) Find the fractional power transmitted in one round trip for the block of material inside a FP-cavity with 2 plane mirrors that have power reflectivity of 0.9. T_{Total} = T^4 frR_m^2 = (5/9)^4 times 0.9^2 =0.07716 b) Find the fractional power loss per round trip. F_l = 1 - T_{Total} = 0.9228 Extra-credit If the block of material has attenuation coefficient alpha = 50 (m^(-1)), now what is the fractional power loss per round trip? F_l = 1 - T_{Total} times e^{-50 times 2 times 0.01} = 0.9716

Nov. 20, 2017
 Consider the life time for nonradiative emission is 1 ns and the life time for radiative emission emission is 1 ns for transition from state 2 to state 1. a) What is the life time of excited atoms in state 2? 1 /(tau_("total")) = 1/(tau_r) + 1/(tau_(rn)) => 1/(tau_("total")) = 2/(10^(-9)) Hence, tau_("total")= 0.5 (ns). b) Assume number atoms in state 2 is 10^9. What is the rate of change in number of atoms in state 2? rate = N / tau = 10^9/(0.5 times 10^{-9}} = 2 times 10^(18) c) Consider a 0.5 m F-P cavity, one mirror has 100% reflectivity, another mirror has 80% reflectivity. If the cavity is filled with a gain medium with gain coefficient of 0.1 cm^{-1}, now an input optical beam with intensity of 1 W/{cm^2} at the 100% mirror, what is the intensity of the beam after 1 round trip? Note that gamma = 0.1 cm^{-1} = 100 times 0.1 m^{-1} I = frR I_o e^(10 times 2 times 0.5} = 0.8 e^(10} = 17.6 ((kW) /(cm^2)) Extra-credit For c), what is the intensity leaving the cavity after 1 round trip? Note that the beam only travel one pass before exiting I_(out) = (1- frR) I_o e^(10 times 0.5} = 0.2 e^(5) = 29.68 ((W) /(cm^2))

Nov. 27, 2017
 Consider an isosceles triangle shaped line shape function in energy (ev) scale. Redraw the diagram in nu (Hz). E_o=1 (ev) => nu_o=c_o/(1.24 times 10 ^(-6)) = 241.9 (THz). Delta E = 0.001 (ev) => Delta nu = 0.01 times 241.9 = 2.419 (THz). Extra-credit Find the value of g_o with the definition of g( nu ), int_0^oo g (nu) d nu=1. Also specify the unit of g_o. (g_o Delta nu )/2 = 1 => g_o = 2/(Delta nu) = 8.268 times 10^(-13) (s), i.e. g_o=0.8268 (ps).

Dec. 4, 2017
 Write a reate equation for level 3. (d N_3)/(dt) = R_3 - (N_3)/(tau_(30)) - (N_3)/(tau_(31)) Extra-credit Write a rate equation for level 1. (d N_1)/(dt) = - (N_1)/(tau_(1)) + (N_3)/(tau_(31)) + W_i N_2 - W_i N_1