Ch-26, Optical properties of materials and materials for opto electronic devices

example-26.1 page no-769

In [3]:
from __future__ import division
#given
#wavelength of light
lamda=5.893*10**-7  #m
#plank's constant
h=6.626*10**-34  #J s
#velocity of light 
c=3*10**8  #m/s
#energy of photon
Ephoton=h*c/lamda  #J
#we know that 1eV=1.6*10**-19 J
#so
EPhoton=Ephoton/(1.6*10**-19)  #eV
print " the energy of photon is %.23f J or %f eV"%(Ephoton,EPhoton)

 the energy of photon is 0.00000000000000000033732 J or 2.108222 eV

example-26.2 page no-769

In [4]:
from math import exp
#given
##thickness of sample
t=0.45*10**-4  #cm
#energy of light
E1=3  #eV
#absorption coefficient
alpha=50000  #per cm
#incident power on the sample
I0=15*10**-3  #W
#we know that
#intensity of transmitted light is given by
It=I0*exp(-alpha*t)  #W or J/s
#thus total energy absorbed is 
Iabsorbed=I0-It  #W or J/s
print " total energy absorbed is %0.3e J/s"%(Iabsorbed)
#plank's constant
h=6.626*10**-34  #J s
#energy of outgoing radiation
E2=2.35  #eV
#fraction of each photon energy unit which is converted ton heat
E=(E1-E2)/E1  
#therefore total amount of energy converted to heat per second is
EC=E*Iabsorbed  #J/s
print " total amount of energy coverted to heat is %0.3e J/s"%(EC)
#charge on an electron
e=1.6*10**-19  #C
#no of photons = nphoton 
nphoton=Iabsorbed/(e*E1)  #photons/sec  (calculation mistake is there in book)
print " the no of photon given off from recombination is %0.3e photons/sec"%(nphoton)

 total energy absorbed is 1.342e-02 J/s
 total amount of energy coverted to heat is 2.907e-03 J/s
 the no of photon given off from recombination is 2.796e+16 photons/sec

example-26.3 page no-783

In [5]:
#given
#energy of photon
E=1.5*10**-19  #J
#quantum eficiency
muquantam=0.6
#photon current
Iopc=3*10**-6  #A
#speed of light
c=3*10**8  #m/s
#plank's constant
h=6.626*10**-34  #J s
#wavelength at which the photodiode is operatng
lamda=h*c/E*10**6  #micro m
#responsivity of diode
R=0.64
#incident optical power is given by
Piop=Iopc/R*10**6  #micro W
print "wavelength at which photodiode is operating is %0.3f micro m \n and incident optical power is %0.3f micro W"%(lamda,Piop)

wavelength at which photodiode is operating is 1.325 micro m 
 and incident optical power is 4.688 micro W

example-26.4 page no-784

In [6]:
from math import asin, pi
#given
#refractive index of core and cladding
mucladding=1.47
mucore=1.50
#critocal angle at the core cladding interface
thetac=asin(mucladding/mucore) *180/(pi) #degrees
print "the critical angle at core cladding interface is %0.3f degrees"%(thetac)

the critical angle at core cladding interface is 78.522 degrees

example-26.5 page no-784

In [7]:
#given
#energy band gap
Eg=0.75*1.6*10**-19  #J
#plank's constant
h=6.626*10**-34  #Js
#speed of light
c=3*10**8  #m/s
#wavelength of light
lamda=h*c/Eg*10**10  #A
print " wavelength of light is %.f A"%(lamda)

 wavelength of light is 16565 A

example-26.7 page no-784

In [8]:
#given
#frequency of light
f=1.5*10**9*10**6  #Hz
#pank's constant
h=6.626*10**-34  #J s
#threshold frequency is
f0=1.2*10**9*10**6  #Hz
#maximum energy of emitted photoelectron is
Emax=h*(f-f0)/(1.6*10**-19)  #eV
print "The maximum enery of the emitted photoe lectron is %0.3f eV"%(Emax)

The maximum enery of the emitted photoe lectron is 1.242 eV