Chapter 10:Optical Properties of Materials

Example 10.1,Page No:10.25

In [1]:
import math

#variable declaration
E2     = 5.56*10**-19;          # Higher Energy level in J
E1     = 2.36*10**-19;          # Lower Energy level in J
h      = 6.626*10**-34;         # plancks constant in J.s
c      = 3*10**8;               # velocity of light in m

# Calculations
dE    = E2 - E1;              # Energy difference in J
lamda = (h*c)/float(dE);             # wavelength in m
 

# Result

print'Wavelength of the photon = %d'%(lamda*10**10),'Å';
print' The colour of the photon is red';

Wavelength of the photon = 6211 Å
 The colour of the photon is red

Example 10.2,Page No:10.25

In [2]:
import math

# Variable declaration
h   = 6.63*10**-34;         # plancks constant in J.s
c   = 3*10**8;              # velocity of light in m
E   = 5.6;                  # bandgap in eV
e   = 1.6*10**-19;           # charge of electron coulombs

# Calculations
lamda    = (h*c)/float(E*e)         # wavelength in m

#output
print'Maximum Wavelength for which diamond is opaque is Imax = %d '%(lamda*10**10),'Å';
print'\n Note: Imax is wrongly printed as 220 Å in textbook';

Maximum Wavelength for which diamond is opaque is Imax = 2219  Å

 Note: Imax is wrongly printed as 220 Å in textbook

Example 10.3,Page No:10.26

In [3]:
import math

#variable declaration
h     = 6.63*10**-34;            # plancks constant
c     = 3*10**8;                 # velocity of light
lamda = 0.6*10**-6;             # wavelength in m
e     = 1.6*10**-19;              # charge of electron
EGap  = 2.25;                    # energy in eV
EGas  = 1.42;                    # energy in eV

#Calculations
E        = (h*c)/float(lamda*e);            # Energy in eV
p_change = (EGap - EGas)/float(100);        #rate of energy gap
x        = (E-EGas)/float(p_change);        #mol % of GaP to be added to get an energy gap of E

# Result
print'Energy of radiation = %3.4f'%E,'eV';
print'Rate of energy gap varies with addition of GaP is %3.5f'%p_change,'eV/mol %';
print'mol percent to be added to get an energy gap of %3.4f'%E,'eV','is %3.2f'%x,'mol %';

Energy of radiation = 2.0719 eV
Rate of energy gap varies with addition of GaP is 0.00830 eV/mol %
mol percent to be added to get an energy gap of 2.0719 eV is 78.54 mol %

Example 10.4,Page No:10.26

In [4]:
import math

#variable declaration
h     = 6.63*10**-34;          #plancks constant in J.s
c     = 3*10**8;               # velocity of light in m
lamda = 1.1*10**-6;             # wavelength in m
e     = 1.6*10**-19;           # charge of electron in coulombs
E2    = 0.4*10**-19;           # energy level in joules


#Calculations
E3  = E2 + ((h*c)/float(lamda));          #energy in J

#Result
print'Energy of the metastable state E3 = %3.1e'%E3,'J';

Energy of the metastable state E3 = 2.2e-19 J

Example 10.5,Page No:10.26

In [5]:
import math

#variable declaration
c   = 3*10**8;               # velocity of light in m
L   = 1.5;                   #length in m
n   = 1.0204;               # refractive index 
BW  = 1.5*10**9;             # Bandwidth in Hz

# Calculations
dV  = c/float(2*L*n);             #frequency in Hz
N   = BW/float(dV);               # Number of optical nodes

# Result
print'Number of Optical modes = % d'%N;

Number of Optical modes =  15

Example 10.6,Page No:10.31

In [6]:
import math

#variable declaration
n1  = 1.55;         # refractive index of core
n2  = 1.53;         # refractive index of cladding


# Calculations
NA  = math.sqrt(n1**2 - n2**2);


#Result
print'Numerical aperture = %3.3f'%NA;

Numerical aperture = 0.248

Example 10.7,Page No:10.31

In [7]:
import math

#variable declaration
n1      = 1.33;         #refractive index of water
n2      = 1;            # refractive index of air

# Calculations
theta_c = math.asin((n2/n1))
theta_c_deg = theta_c*(180/float(math.pi));        # radian to degree conversion

# Result
print'For angles above %3.2f° ,there will be total internal reflection in water'%theta_c_deg ;

For angles above 48.75° ,there will be total internal reflection in water