Chapter 1: Interference of Light

Example 1.2.1, Page number 1-11

In [1]:
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 [33]:
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):
    lamda = (4*u*t*int(math.cos(math.radians(90))))/((2*n)-1)
    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

Please note: Since r=90, cos(r)=0
Hence, the answers given in the textbook are incorrect

Example 1.2.3, Page number 1-12

In [3]:
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 [4]:
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 [5]:
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 [6]:
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 [7]:
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 [2]:
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 [5]:
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 [14]:
#Variable declaration 
a = 40.       #angle(sec)
lamda = 1.2  #distance between fringes(cm)
alpha = 10   #no. of fringes

#Calculations
Bair = lamda/alpha          #cm
alpha = (a*math.pi)/(3600*180)   #radians
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 [16]:
#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 [18]:
#Variable declaration 
u = 1.4              #refractive index
B = 0.25             #fringe spacing(cm)
a = 20               #angle(secs)

#Calculations
alpha = (a*math.pi)/(3600*180)  #radians
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 [20]:
#Variable declaration 
u = 1.5              #refractive index
lamda = 5.82*10**-5  #wavelength(cm)
a = 20               #angle(secs)

#Calculations
alpha = (a*math.pi)/(3600*180)  #radians
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 [23]:
#Variable declaration 
u = 1                #refractive index for air film
lamda = 6*10**-5     #wavelength(cm)
B = 1./10            #distance between fringes(cm)

#Calculations
alpha = lamda/(2*u*B)  #radians
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 [26]:
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 [27]:
#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 [28]:
#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
print "Radius of curvature =",R,"cm"

Radius of curvature = 125.0 cm

Example 1.4.3, Page number 1-33

In [34]:
#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
print "Radius of curvature =",round(R,2),"cm"

Radius of curvature = 99.83 cm

Example 1.4.4, Page number 1-33

In [36]:
#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 [37]:
#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 [35]:
#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 [42]:
#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 [43]:
#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 [46]:
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 [50]:
#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 [55]:
#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 [56]:
#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