ECE 240a Midterm Exam
1. Wavefunctions (20 pts.)
a) The diagram below shows an asymmetric potential well such as the one presented in HW 1
Sketch the first three allowed wavefunctions n (x) for n = 1, 2, 3.
3(x)
1(x)
V2
V1
2(x)
b) Now consider the three f
ECE 240a Fall 2016
Problem Set 1 Solutions
1. Index of refraction and complex susceptibility
The (complex) index of refraction is defined by
.
n2c () = ()/0 = r () = 1 + ()
(1)
where r () is the relative permitivity, 0 is the permitivity of free space, an
ECE 240a Fall 2016 Problem Set 3
Due: 11/8 in class
1) The time-harmonic complex electric field is given by
b
Ey (x, z) = Ey (x)eiz y
(1)
a) Using Maxwells equations in time-harmoninc form, determine an expression for the
magnetic field
b)Determine an exp
ECE 240a Fall 2016
Problem Set 2 - due Thursday Oct. 20
1. Eigenstates and eigenvalues of a quantized harmonic oscillator
The potential function for a quantized harmonic oscillator is given by
V (x) =
1
1
Kx2 = m 2 x2 .
2
2
a) Show that the Schrdinger equ
ECE 240a Fall 2016
Problem Set 2 Solutions
1. Eigenstates and eigenvalues of a quantized harmonic oscillator
The potential function for a quantized harmonic oscillator is given by
V (x) =
1
1
Kx2 = m 2 x2 .
2
2
a) Show that the Schrdinger equation for thi
ECE 240a - Notes on Spontaneous
Emission within a Cavity
1 Introduction
Many treatments of lasers treat the rate of spontaneous emission as specified by the time constant sp
as a constant that is independent of the effect of the cavity. In this set of not
ECE 240a Fall 2016
Problem Set 3 Solutions
1. The time-harmonic complex electric field is given by
b
Ey (x, z) = Ey (x)eiz y
(1)
a) Using Maxwells equations in time-harmonic form, determine an expression
for the magnetic field
Solution
Using Maxwells equa
ECE 240 Take Home Final Design Project - Fall 2016
Due: Dec. 9 at noon in room 3402
This problem considers the design of a diode-pumped green laser pointer with a general configuration shown below.
The goal is to create the highest output CW power at the
ECE 240a Fall 2016
Problem Set 1
Due: 10/11 in class
1. Index of refraction and complex susceptibility
The (complex) index of refraction is defined by
.
n2c () = ()/0 = r () = 1 + ()
(1)
where r () is the relative permitivity, 0 is the permitivity of free
ECE 240a Midterm Exam Solutions
1. Upconversion Laser (60 pts.)
It is possible to have the lasing photon energy greater than the pump photon energy for a
material system shown in the gure below. This kind of system is called an upconversion laser.
Pump (W
ECE 240a
Problem Set 2 Solutions
1. Eigenstates and eigenvalues of a quantized harmonic oscillator
The potential function for a quantized harmonic oscillator is given by
1
1
Kx2 = m 2 x2 .
2
2
V (x) =
a) Show that the Schrdinger equation for this potentia
ECE 240a
Problem Set 3 Solutions
1. The time-harmonic complex electric eld is given by
Ey (x, z ) = Ey (x)eiz y
(1)
a) Using Maxwells equations in time-harmoninc form, determine an expression
for the magnetic eld
Solution
Using Maxwells equations, we can
ECE 240a
Problem Set 4 Solutions
1. Verdeyen 5.6
The resonant frequencies are given by (6.5.4)
m,p,q =
(1 + m + p)
c
q+
cos1 (g1 g2 )1/2
2nd
where
g1,2 = 1
d
R1,2
R (z ) = z 1 +
z0
z
and
2
.
The mirror radius of curvature is
z0
R1 = z 1 +
z
2
= 25 1 +
12
ECE 240a Midterm Exam Solutions
Fall 2011
1. Assume that for a one-dimensional, two-state material system, the solutions to the
Schrdinger equation in the absence of an optical eld are 1 (x) = cos(kx) and
2 (x) = sin(kx). Both functions are dened over a r
ECE 240A Lasers and Optics"
Homework 3
Problem 1: Surface plasmon waveguide
The interface between a metal (n3) and air (n1=1) can support the propagation of a surface wave, known as
a surface plasmon, provided that the metal has a negative dielectric
Problem1:Gaussianbeams
(a)wa=wbandRa=Rb.
(b)1exp(2a2/w2)whichevaluatesto86.5%
Problem2:ABCDMatrices
n 1 d
d
1
n R1
n
(a)
n 12 d
1
1
n 1 d
(n 1)( ) 1
R1 R2
n R2
n R1 R2
(b)w0>0(actually,/2n,thelimitoffocusing)
(c)Thiswouldnotmakeagoodresonatorsincew(
University of California, San Diego
Department of Electrical and Computer Engineering
ECE 240A Optics and Lasers Fall 2007-Final Examination
Monday December 10, 2007. Time limit: 2 hours
Answeronlyfourofthefollowingsixproblems(ifyouanswermore,thefirstfour
ECE 240a Fall 2013
Problem Set 1 Solutions
1. Index of refraction and complex susceptibility
The (complex) index of refraction is dened by
.
n2 ( ) = ( )/0 = r ( ) = 1 + ( )
c
(1)
where r ( ) is the relative permitivity, 0 is the permitivity of free space