# Chapter 1: Interference of Light¶

## Example 1.2.1, Page number 1-11¶

In :
import math

#Variable declaration
i = 45        #angle of incidence(degrees)
t = 4*10**-5  #thickness of film(cm)
u = 1.2

#Calculations & Result
r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))
for n in range(1,4):
lamda = (2*u*t*math.cos(r*math.pi/180))/n
print "For n = %d, wavelength = %.2f A"%(n,lamda/10**-8)
print "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range"

For n = 1, wavelength = 7756.29 A
For n = 2, wavelength = 3878.14 A
For n = 3, wavelength = 2585.43 A
Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths lie in this range


## Example 1.2.2, Page number 1-12¶

In :
import math

#Variable declaration
r = 90        #angle of refraction(degrees)
t = 5*10**-5  #thickness of film(cm)
u = 1.33

#Calculations & Result
for n in range(1,4):
print "For n = %d, wavelength = %.2f A"%(n,lamda)
print "Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range"

print "\nPlease note: Since r=90, cos(r)=0\nHence, the answers given in the textbook are incorrect"

For n = 1, wavelength = 0.00 A
For n = 2, wavelength = 0.00 A
For n = 3, wavelength = 0.00 A
Since the visible range of wavelengths lie between 4000 to 7000 A, none of the obtained wavelengths do not lie in this range

Hence, the answers given in the textbook are incorrect


## Example 1.2.3, Page number 1-12¶

In :
import math

#Variable declaration
i = 45           #angle of incidence(degrees)
t = 1.5*10**-4   #thickness of film(cm)
lamda = 5*10**-5 #wavelength(cm)
u = 4./3.        #refractive index

#Calculations
r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))
n = (2*u*t*math.cos(r*math.pi/180))/lamda

#Result
print "The order of interfernce is %.2f, close to 7"%n

The order of interfernce is 6.78, close to 7


## Example 12.2.4, Pae number 1-13¶

In :
import math

#Variable declaration
i = 45              #angle of incidence(degrees)
lamda = 5896*10**-8 #wavelength(cm)
u = 1.33            #refractive index
n = 1

#Calculations
r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))
t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)

#Result
print "The required thickness is",round((t/1E-5),2),"*10^-5 cm"

The required thickness is 1.31 *10^-5 cm


## Example 1.2.5, Page number 1-14¶

In :
lamda1 = 7000  #wavelength(A)
lamda2 = 5000  #wavelength(A)
u = 1.3        #R.I. of oil

#Calculations
'''
2utcosr = (2n-1)7000/2   ----(1)
2utcosr = (2n+1)5000/2   ----(2)
Divinding (1) by (2), we get the following expression
1 = (2n+1)5000
-----------
(2n-1)7000
Solving the above expression, we get,
'''
n = 12000/4000
t = ((2*n-1)*lamda)/(2*u*math.cos(r*math.pi/180)*2)

#Result
print "The required thickness is",round((t/1E-5),4),"*10^-5 cm"

The required thickness is 6.6936 *10^-5 cm


## Example 1.2.6, Page number 1-15¶

In :
import math

#Variable declaration
i = 30              #angle of incidence(degrees)
lamda = 5890*10**-8 #wavelength(cm)
u = 1.46            #refractive index
n = 8

#Calculations
r = math.degrees(math.asin(math.sin(i*math.pi/180)/u))
t = (n*lamda)/(2*u*math.cos(r*math.pi/180))

#Result
print "The required thickness is",round((t/1E-4),3),"*10^-4 cm"

The required thickness is 1.718 *10^-4 cm


## Example 1.2.7, Page number 1-15¶

In :
import math

#Variable declaration
r = 60              #angle of refraction(degrees)
lamda = 5890*10**-8 #wavelength(cm)
u = 1.5             #refractive index
n = 1               #for minimumm thickness

#Calculations
#For r = 60
t1 = (n*lamda)/(2*u*math.cos(r*math.pi/180))

#For normal incidence
r = 0
t2 = (n*lamda)/(2*u*math.cos(r*math.pi/180))

#Result
print "For r = 60, the required thickness is",round((t1/1E-5),2),"*10^-5 cm"
print "For r = 0, the required thickness is",round((t2/1E-5),2),"*10^-5 cm"

For r = 60, the required thickness is 3.93 *10^-5 cm
For r = 0, the required thickness is 1.96 *10^-5 cm


## Example 1.2.8, Page number 1-16¶

In :
import math

#Variable declaration
r = 0               #for normal incidence(degrees)
lamda = 5.5*10**-5  #wavelength(cm)
n = 1               #for minimumm thickness
A = 10**4           #area(cm^2)
V = 0.2             #volume(cc)

#Calculations
t = V/A
#for nth dark band,
u = (n*lamda)/(2*t*math.cos(r*math.pi/180))

#Result
print "Refractive index =",u

Refractive index = 1.375


## Example 1.2.9, Page number 1-17¶

In :
import math

#Variable declaration
r = 60        #angle of incidence(degrees)
t = 2*10**-7  #thickness of film(cm)
u = 1.2

#Calculations & Result
for n in range(1,4):
lamda = (4*u*t*math.cos(r*math.pi/180))/(2*n-1)
print "For n = %d, wavelength = %.2f A"%(n,lamda/10**-10)
print "Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A"

For n = 1, wavelength = 4800.00 A
For n = 2, wavelength = 1600.00 A
For n = 3, wavelength = 960.00 A
Since the visible range of wavelengths lie between 4000 to 7000 A, the wavelength in the visible spectrum is 4800 A


## Example 1.3.1, Page number 1-21¶

In :
#Variable declaration
a = 40.       #angle(sec)
lamda = 1.2  #distance between fringes(cm)
alpha = 10   #no. of fringes

#Calculations
Bair = lamda/alpha          #cm
lamda = 2*alpha*Bair

#Result
print "Wavelength of monochromatic light =",round((lamda/1E-8),1),"A"

Wavelength of monochromatic light = 4654.2 A


## Example 1.3.2, Page number 1-22¶

In :
#Variable declaration
lamda = 5893*10**-8   #wavelength(cm)
u = 1.52              #refractive index
B = 0.1               #fringe spacing(cm)

#Calculations
alpha = (lamda/(2*u*B))*180*3600/math.pi   #seconds

#Result
print "Angle of wedge =",round(alpha,2),"secs"

Angle of wedge = 39.98 secs


## Example 1.3.3, Page number 1-22¶

In :
#Variable declaration
u = 1.4              #refractive index
B = 0.25             #fringe spacing(cm)
a = 20               #angle(secs)

#Calculations
lamda = 2*u*alpha*B

#Result
print "Wavelength of light =",round((lamda/1E-8),2),"A"

Wavelength of light = 6787.39 A


## Example 1.3.4, Page number 1-23¶

In :
#Variable declaration
u = 1.5              #refractive index
lamda = 5.82*10**-5  #wavelength(cm)
a = 20               #angle(secs)

#Calculations
B = lamda/(2*u*alpha)
N = 1/B

#Result
print "Number of interfernce fronges pr cm is",round(N)

Number of interfernce fronges pr cm is 5.0


## Example 1.3.5, Page number 1-24¶

In :
#Variable declaration
u = 1                #refractive index for air film
lamda = 6*10**-5     #wavelength(cm)
B = 1./10            #distance between fringes(cm)

#Calculations
d = alpha*10

#Result
print "Daimeter of wire =",d,"cm"

Daimeter of wire = 0.003 cm


## Example 1.3.6, Page number 1-24¶

In :
alpha = 0.01*10**-1/10   #angle(radians)
u = 1                    #refractive index for air film
lamda = 5900*10**-10     #wavlength(m)

#Calculation
B = lamda/(2*u*alpha)

#Result
print "Seperation between fringes is",B/10**-3,"mm"

Seperation between fringes is 2.95 mm


## Example 1.4.1, Page number 1-32¶

In :
#Variable declaration
n = 40

#Calculation
#Equating the equation 4*R*n*lamda=4*4R*n*lamda, we get

N = (4*4*n)/4

#result
print "Ring number =",N

Ring number = 160


## Example 1.4.2, Page number 1-32¶

In :
#Variable declaration
n = 10
Dn = 0.5         #diameter of dark ring(cm)
lamda = 5*10**-5 #waelength(cm)

#Calculations
R = Dn**2/(4*n*lamda)

#Result

Radius of curvature = 125.0 cm


## Example 1.4.3, Page number 1-33¶

In :
#Variable declaration
n = 5
p = 10
D5 = 0.336          #diameter of 5th ring(cm)
lamda = 5890*10**-8 #waelength(cm)
D15 = 0.59          #diameter of 15th ringcm

#Calculations
R = (D15**2-D5**2)/(4*p*lamda)

#Result

Radius of curvature = 99.83 cm


## Example 1.4.4, Page number 1-33¶

In :
#Varaible declaration
Dn = 0.42     #diameter of dark ring(cm)
p = 8
R = 200       #radius of curvature(cm)
Dn8 = 0.7     #diameter of (n+8)th ring(cm)

#Calculations
lamda = (Dn8**2-Dn**2)/(4*R*p)

#Result
print "Wavelength =",lamda/1E-8,"A"

Wavelength = 4900.0 A


## Example 1.4.5, Page number 1-34¶

In :
#Varaible declaration
Dn = 0.218             #cm
Dn10 = 0.451           #cm
lamda = 5893*10**-8    #wavelength(cm)
R = 90                 #radius of curvature(cm)
p = 10

#Calculation
u = (4*p*lamda*R)/(Dn10**2-Dn**2)

#Result
print "Refractive index =",round(u,3)

Refractive index = 1.361


## Example 1.4.6, Page number 1-34¶

In :
#Varaible declaration
D5 = 0.42     #diameter of dark ring(cm)

#Calculations
'''
For 5th dark ring,
D5^2 = 20*R*lamda  -----1

For 10th dark ring,
D10^2 = 40*R*lamda -----2

Substituting 1 in 2,
'''

D10 = math.sqrt((40*D5**2)/20)

#Result
print "Diameter of the 10th dark ring =",round(D10,3),"cm"

Diameter of the 10th dark ring = 0.594 cm


## Example 1.4.7, Page number 1-35¶

In :
#Varaible declaration
lamda_n = 6000    #wavelength of nth ring(A)
lamda_n1 = 5000   #wavelength for (n+1)th ring(cm)

#Calculations
'''
Dn^2 = 4*R*n*lamda_n ---1

Dn+1^2 = 4*R(n+1)*lamda_n1  ---2

Equating 1 and 2, we get,
'''

n = 5
R = 2
Dn = math.sqrt(4*R*n*lamda_n*10**-8)

#Result
print "Diameter =",round(Dn,3),"cm"

Diameter = 0.049 cm


## Example 1.4.8, Page number 1-35¶

In :
#Varaible declaration
Dair = 2.3   #diameter of ring in air(cm)
Dliq = 2     #diameter of ring in liquid(cm)

#Calculations
u = Dair**2/Dliq**2

#Result
print "Refractive index =",u

Refractive index = 1.3225


## Example 1.4.11, Page number 1-37¶

In :
D4 = 0.4    #diameter of 4th ring(cm)
D12 = 0.7   #diameter of 12th ring(cm)

#Calculations
'''
For D4,
D4 = math.sqrt(4R*4*lamda)
'''
rt_Rl = 0.1
R = 80

#For D20,
D20 = math.sqrt(R)*rt_Rl

#Result
print "Diameter of 20th ring =",round(D20,3),"cm"

Diameter of 20th ring = 0.894 cm


## Example 1.4.12, Page number 1-36¶

In :
#Variable declaration
n = 5
p = 10
D5 = 0.336          #diameter of 5th ring(cm)
D15 = 0.590         #diameter of 15th ring(cm)
R = 100             #radius of curvature(cm)

#Calculations
lamda = (D15**2-D5**2)/(4*R*p)

#Result
print "Wavlength =",lamda/1E-8,"A"

Wavlength = 5880.1 A


## Example 1.7.1, Page number 1-42¶

In :
#Variable declaration
lamda = 560 #wavelength(nm)
u = 2       #refractive index

#Calculations
lamda_dash = lamda/u
t = lamda_dash/4

#Result
print "Thickness of film =",t,"nm"

Thickness of film = 70 nm


## Example 1.7.2, Page number 1-42¶

In :
#Variable declaration
lamda = 6000 #wavelength(E)
u = 1.2       #refractive index

#Calculations
lamda_dash = lamda/u
t = lamda_dash/4

#Result
print "Thickness of film =",t,"A"

Thickness of film = 1250.0 A