# Chapter 5 - Polarisation¶

## Example 4 - pg 134¶

In [1]:
#pg 134
#calculate the angle required
#Given:
import math
mu = 1.33; #Refractive index of water
#Brewster's angle, theta_p = atand(mu) ;
#calculations
theta_p = math.atan(mu)*57.3; # in degrees
theta_s = 90-theta_p ; # in degrees
#results
print "Angle (degrees) = ",round(theta_s,1)

Angle (degrees) =  36.9


## Example 5 - pg 136¶

In [3]:
#pg 136
#calculate the Critical angle
import math
#Given:
r = 90.; # in degrees
mu_o= 1.658 ;# Refractive index for ordinary array
mu =1.55; # Refractive index for a canada balsam material
#calculations
#Snell's Law,mu1*sin(i) = mu2*sin(r), we have :
i = math.asin((mu*math.sin(r/57.3))/mu_o)*57.3; # angle in degrees
#results
print "Critical angle (degrees) = ",round(i,0)

Critical angle (degrees) =  69.0


## Example 6 - pg 147¶

In [5]:
#pg 147
#calculate the Minimum thickness of Plane and Circularly polarised light
#Given :
mu_o = 1.544; #Refractive index for ordinary ray
mu_e = 1.553;#Refractive index for extraordinary ray
lambd = 5890.;#Wavelength in A
#calculations
#(a)Plane polarised light :
#lambd is converted from A to cm , 1 A = 1.0*10**-8 cm
t1 = (lambd*10**-8)/(2*(mu_e-mu_o));#Minimum thickness in cm
#(b)Circularly polarised light :
t2 = (lambd*10**-8)/(4*(mu_e-mu_o));# Minimum thickness in cm
#results
print"Minimum thickness :"
print"(a)Plane polarised light :",round(t1*1000.,2)," x 10**-3 cm "
print"(b)Circularly polarised light :",round(t2*1000.,2),"x 10**-3 cm "

Minimum thickness :
(a)Plane polarised light : 3.27  x 10**-3 cm
(b)Circularly polarised light : 1.64 x 10**-3 cm


## Example 7 - pg 150¶

In [7]:
#pg 150
#calculate the phase difference
#Given :
import math
lambd = 5890.; #Wavelength in A
#(a)Calcite crystal
mu1_o = 1.658;#refractive index for ordinary ray
mu1_e = 1.486;#refractive index for extraordinary ray
t1 = 0.0052 ; #thickness in mm
# 1 A = 1.0*10**-7 mm
alpha1 = ((2*math.pi*(mu1_o-mu1_e)*t1)/(lambd*10**-7)); # phase difference in radians
#(b) Quartz crystal
mu2_o = 1.544; #refractive index for ordinary ray
mu2_e = 1.553; #refractive index for extraordinary ray
t2 = 0.0234;#thickness in mm
alpha2 = ((2*math.pi*(mu2_e-mu2_o)*t2)/(lambd*10**-7)); # phase difference in radians
print"(a)Calcite crystal : \n  Phase difference is (radians) = ",round(alpha1,3)
print"(a)Quartz crystal : \n  Phase difference is (radians) = ",round(alpha2,3)

(a)Calcite crystal :
Phase difference is (radians) =  9.541
(a)Quartz crystal :
Phase difference is (radians) =  2.247


## Example 9 - pg 159¶

In [8]:
#pg 159
#calculate the concentration and Change in angle
#Given :
rho = 6.6; # Specific rotation of sugar in degrees g^-1 cm^2
l = 20; #length in cm
deltad = 1*10**-3;#difference in sugar concentration in g/cm^3
lc = 0.1; # least count in degrees
#Rotation due to optical activity  = rho*l*d
#calculations
deltatheta = rho*l*deltad; # in degrees
#results
print"Change in theta (degrees) = ",deltatheta

if(deltatheta > lc):
print"The concentration of 1 mg/cm^3 will be detected by the given urinalysis tube."
else:
print"The concentration of 1 mg/cm^3 will not be detected."

Change in theta (degrees) =  0.132
The concentration of 1 mg/cm^3 will be detected by the given urinalysis tube.