Chapter 1: Interference

Example 1.1, Page number 1-16

In [1]:
import math

#Given Data:
i=45*math.pi/180         #angle of incidence
u=1.33                   #Refractive index of a soap film
lamda=5.896*10**-7       #wavelength of required yellow light

#Calculations:
#u=sin i/sin r           #Snell's law   .So,
r=math.asin(math.sin(i)/u)     #angle of reflection

#Now, condition for bright fringe is
#2ut*cos r=(2n-1)lamda/2
#Here n=1
t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film at which light will appear bright yellow
print"Minimum thickness of film at which light will appear bright yellow of required wavelength is =",t,"m"

Minimum thickness of film at which light will appear bright yellow of required wavelength is = 1.30853030399e-07 m

Example 1.2, Page number 1-16

In [4]:
import math

#Given Data:
theta=40./3600*math.pi/180     #angle of wedge in radians
B=0.12*10**-2                 #fringe spacing

#Calculations:
#We know, B=lam/(2*u*theta). Here u=1
lamda=2*B*theta               #wavelength of light used
print"Wavelength of light used is =",lamda,"m"

Wavelength of light used is = 4.65421133865e-07 m

Example 1.3, Page number 1-17

In [6]:
import math

#Given Data:
i=30*math.pi/180        #angle of incidence
u=1.46                  #Refractive index of a oil
lamda=5.890*10**-7       #wavelength of required yellow light
n=8                     #eighth dark band

#Calculations:
#u=sin i/sin r      #Snell's law   .So,
r=math.asin(math.sin(i)/u)     #angle of reflection

#Now, condition for dark fringe is
#2ut*cos r=n*lamda
t=n*lamda/(2*u*math.cos(r))     #thickness of film
print"Thickness of the film is =",t,"m"

Thickness of the film is = 1.71755887917e-06 m

Example 1.4, Page number 1-17

In [7]:
import math

#Given Data:

B=0.1*10**-2               #fringe spacing
lamda=5.893*10**-7         #Wavelength of light
u=1.52                     #Refractive index of wedge

#Calculations:
#We know, B=lamda/(2*u*theta). Here u=1
theta1=lamda/(2*u*B)            #angle of wedge in radians
theta=theta1*3600*180/math.pi   #angle of wedge in seconds
print"Angle of wedge is =",theta,"seconds of an arc"

Angle of wedge is = 39.9841612899 seconds of an arc

Example 1.5, Page number 1-18

In [9]:
import math

#Given Data:
t=0.2/(100)**2*10**-2     #thickness of film in meter
lamda=5.5*10**-7         #wavelength of light in meter
r=0                      #normal incidence
n=1                      #first band

#Calculations:

#Condition for dark fringe is
#2ut*cos r =n*lamda
u=n*lamda/(2*t*math.cos(r))   #Refractive index of a oil
print"Refractive index of a oil is =",u

Refractive index of a oil is = 1.375

Example 1.6, Page number 1-18

In [4]:
import math

#Given Data:
lamda=5.893*10**-7    #Wavelength of light
u=1.42                #Refractive index of a soap film
r=0                   #normal incidence
n=1                   #first band

#Calculations:

#(i)
#Condition for dark fringe is
#2ut*cos r=n*lamda
t1=n*lamda/(2*u*cos(r))      #thickness of film for dark black fringe
print"Thickness of the film for dark black fringe is =",t1,"m"

#(ii)
#Now, condition for bright fringe is
#2ut*cos r=(2n-1)lamda/2
t2=lamda/(2*2*u*math.cos(r))      #Thickness of film for bright fringe
print"Thickness of film for bright fringe is =",t2,"m"

Thickness of the film for dark black fringe is = 2.075e-07 m
Thickness of film for bright fringe is = 1.0375e-07 m

Example 1.7, Page number 1-19

In [49]:
import math 
#Given Data:
i=30*math.pi/180         #angle of incidence
u=1.43               #Refractive index of a soap film
lamda=6*10**-7          #wavelength of light
n=1                  #For minimum thickness

#Calculations:
#u=sin i/sin r      #Snell's law   .So,
r=(math.asin(math.sin(i)/u))    #angle of reflection

#Now, condition of minima in transmitted system is
#2ut*cos(r)=(2n-1)lamda/2
t=lamda/(2*2*u*math.cos(r))      #minimum thickness of film
print"Minimum thickness of film is ",t,"m"

Minimum thickness of film is  1.11962124395e-07 m

Example 1.8, Page number 1-19

In [2]:
import math

#Given Data:

lamda = 5893*10**-10      #Wavelength of light
theta = 1               #assuming value of theta

#We know, B=lamda/(2*u*theta). Here u=1
B = lamda/(2*theta)       #fringe spacing
n=20                  #interference fringes

#Calculations:
#t=n*B*tan(theta)
t = 20*B*theta          #Thickness of wire
print"Thickness of wire is =",t,"m"

Thickness of wire is = 5.893e-06 m

Example 1.9, Page number 1-20

In [5]:
import math

#Given Data:
u1=1.3                #Refractive index of oil
u2=1.5                #Refractive index of glass
lamda1=7*10**-7       #Wavelength of light
lamda2=5*10**-7       #Wavelength of light

#Calculations:

#for finding value of n, solve:
#(2n+1)*lamda1/2=(2(n+1)+1)*lamda2/2
#We get,n=2
n=2

toil=(2*n+1)*lamda1/(2*u1*2)     #thickness of oil layer
print"Thickness of oil layer is =",toil,"m"

Thickness of oil layer is = 6.73076923077e-07 m
In [ ]:
Example 1.10, Page number 1-21
In [6]:
import math

#Given Data:
u1=1.2               #Refractive index of drop of oil
u2=1.33              #Refractive index of water
lamda=4.8*10**-7     #wavelength of light
n=3                  #order
r=0                  #normal incidence,so r=0

#Calculations:
t=n*lamda/(2*u1)      #Thickness of oil drop
print"Thickness of oil drop is =",t,"m"

Thickness of oil drop is = 6e-07 m

Example 1.11, Page number 1-22

In [55]:
import math

#Given Data:
i=math.asin(4/5)          #angle of incidence
u=4/3                    #Refractive index of a soap film
lamda1=6.1*10**-7        #wavelength of light
lamda2=6*10**-7          #wavelength of light

#Calculations:
#u=sin i/sin r      #Snell's law   .So,
r=math.asin(math.sin(i)/u)     #angle of reflection

#Now, condition for dark band is
#2ut*cos r=n*lamda
#for consecutive bands, n=lamda2/(lamda1-lamda2). hence

t=lamda2*lamda1/((lamda1-lamda2)*2*u*math.sqrt(1-(math.sin(i)/u)**2))     #thickness of film
print"Thickness of the film is =",t,"m"

Thickness of the film is = 1.83e-05 m

Example 1.12, Page number 1-40

In [9]:
import math

#Given Data:
n=10                #10th dark ring
Dn=0.5*10**-2       #Diameter of ring
lamda=6*10**-7      #wavelength of light

#Calculations:
#As Dn^2=4*n*R*lamda
R=Dn**2/(4*n*lamda)    #Radius of curvature of the lens
print"Radius of curvature of the lens is =",R,"m"

t=Dn**2/(8*R)        #thickness of air  film
print"Thickness of air  film is =",t,"m"

Radius of curvature of the lens is = 1.04166666667 m
Thickness of air  film is = 3e-06 m

Example 1.13, Page number 1-41

In [13]:
import math

#Given Data:

B=0.25*10**-2             #fringe spacing
lamda=5.5*10**-7          #Wavelength of light
u=1.4                     #Refractive index of wedge

#Calculations:
#We know, B=lamda/(2*u*theta).
theta1=lamda/(2*u*B)            #angle of wedge in radians
theta=theta1*3600*180/math.pi   #angle of wedge in seconds
print"Angle of wedge is =",theta,"seconds"

Angle of wedge is = 16.2065204908 seconds

Example 1.14, Page number 1-41

In [14]:
import math

#Given Data:
n=4              #4th dark ring
m=12             #m=n+p
D4=0.4*10**-2    #Diameter of 4th ring
D12=0.7*10**-2   #Diameter of 12th ring

#Calculations:

#(Dn+p)^2-Dn^2=4*p*lamda*R
#Solving, (D12^2-D4^2)/(D20^2-D4^2)
#We get above value =1/2. Hence
D20=math.sqrt(2*D12**2-D4**2)      #Diameter of 20th ring
print"Diameter of 20th ring is =",D20,"m"

Diameter of 20th ring is = 0.00905538513814 m

Example 1.15, Page number 1-42

In [15]:
import math

#Given Data:
n=6                  #6th bright ring
D6=0.31*10**-2       #Diameter of 6th ring
lamda=6*10**-7       #wavelength of light
R=1                  #Radius of curvature

#Calculations:

#Diameter of nth bright ring is 
#Dn^2=2(2n-1)*lamda*R/u. Hence
u=2*(2*n-1)*lamda*R/(D6)**2    #Refractive index of liquid
print"Refractive index of liquid is =",u

Refractive index of liquid is = 1.37356919875

Example 1.16, Page number 1-42

In [16]:
import math

#Given Data:
lamda=6*10**-7         #wavelength of light
k=0.125*10**-4         #k=D(n+1)^2-Dn^2.
u=1                    #Refractive index of medium between lens and plate
#Calculations:

#(i)
lamda1=4.5*10**-7      #new wavelength of light
#Difference between squres of diameters of successive rings is directly proportional to wavelength.So,
k1=lamda1/lamda*k      #new Difference between squres of diameters of successive rings after changing wavelength
print"New Difference between squres of diameters of successive rings after changing wavelength is =",k1,"m^2"

#(ii)
u2=1.33          #Refractive index of liquid introduced between lens and plate
#Difference between squres of diameters of successive rings is inversely proportional to Refractive index.so,
k2=u/u2*k        #new Difference between squres of diameters of successive rings after changing refractive index
print"New Difference between squres of diameters of successive rings after changing refrective index is =",k2,"m^2"

#(iii)
#Difference between squres of diameters of successive rings is directly proportional to Radius of curvature.So,
#after doubling radius of curvature,
k3=2*k           #new Difference between squres of diameters of successive rings after doubling radius of curvature
print"New Difference between squres of diameters of successive rings after doubling radius of curvature is =",k3,"m^2"

New Difference between squres of diameters of successive rings after changing wavelength is = 9.375e-06 m^2
New Difference between squres of diameters of successive rings after changing refrective index is = 9.3984962406e-06 m^2
New Difference between squres of diameters of successive rings after doubling radius of curvature is = 2.5e-05 m^2

Example 1.17, Page number 1-43

In [17]:
import math

#Given Data:
Dn=0.225*10**-2         #Diameter of nth ring
Dm=0.45*10**-2          #Diameter of (n+9)th ring
lamda=6*10**-7          #wavelength of light
R=0.9                   #Radius of curvature
p=9

#Calculations:
#(Dn+p)^2-Dn^2=4*p*lamda*R/u
u=4*p*lamda*R/((Dm)**2-Dn**2)     #Refractive index of liquid
print"Refractive index of liquid is =",u

Refractive index of liquid is = 1.28

Example 1.18, Page number 1-44

In [18]:
import math

#Given Data:
D10=0.5*10**-2          #Diameter of 10th ring
lamda=5.5*10**-7        #wavelength of light
u=1.25                  #Refractive index of liquid


#Calculations:
#As Dn^2=4*n*R*lamda/u
#Dn^2 is inversely proportional to refractive index.
D10n=D10/math.sqrt(u)        #new diameter of 10th ring after changing medium between lens and plate
print"new diameter of 10th ring after changing medium between lens and plate is =",D10n,"m"

new diameter of 10th ring after changing medium between lens and plate is = 0.004472135955 m

Example 1.19, Page number 1-45

In [20]:
import math

#Given Data:
D5=0.336*10**-2       #Diameter of 5th ring
D15=0.59*10**-2       #Diameter of 15th ring
lamda=5.89*10**-7       #wavelength of light
p=10                  #n=5,n+p=15

#Calculations:
#(Dn+p)^2-Dn^2=4*p*lamdaR/u
R=((D15)**2-D5**2)/(4*p*lamda)      #Radius of curvature of the lens
print"Radius of curvature of the lens is =",R,"m"

Radius of curvature of the lens is = 0.998319185059 m

Example 1.20, Page number 1-45

In [27]:
import math

#Given Data:
n=10               #10th dark ring
D10=0.6*10**-2     #Diameter of ring
lamda=6*10**-7     #wavelength of light
u=4./3              #Refractive index of water


#Calculations:
#As Dn^2=4*n*R*lamda/u
R=(D10**2)*u/(4*n*lamda)    #Radius of curvature of the lens
print"Radius of curvature of the lens is =",R,"m"

Radius of curvature of the lens is = 2.0 m

Example 1.20.1, Page number 1-52

In [30]:
import math

#Given Data:
i=45*math.pi/180        #angle of incidence
u=1.2                   #Refractive index of a film
t=4*10**-7              #thickness of film

#Calculations:
#u=sin i/sin r       #Snell's law   .So,
r=math.asin(math.sin(i)/u)     #angle of reflection

#Now, condition for dark fringe is
#2ut*cos r=n*lamda
lamda1=2*u*t*math.cos(r)/1          #n=1
print"For n=1 wavelength is =",lamda1,"m"
print"This is in the visible spectrum and it will remain absent." 

lamda2=2*u*t*math.cos(r)/2          #n=2
print"For n=2 wavelength is =",lamda2,"m"
print"This is not in the visible spectrum"

For n=1 wavelength is = 7.75628777187e-07 m
This is in the visible spectrum and it will remain absent.
For n=2 wavelength is = 3.87814388593e-07 m
This is not in the visible spectrum

Example 1.20.2, Page number 1-53

In [31]:
import math

#Given Data:
r=45*math.pi/180        #angle of refraction
u=1.45                  #Refractive index of a medium
lamda=5.5*10**-7        #wavelength of required yellow light
n=1

#Calculations:

#Now, condition for dark fringe is
#2ut*cos r=n*lamda
t=n*lamda/(2*u*math.cos(r))      #thickness of thin medium
print"Thickness of the thin medium is =",t,"m"

Thickness of the thin medium is = 2.68212917002e-07 m

Example 1.21, Page number 1-45

In [32]:
import math

#Given Data:
u=1.33                 #Refractive index of a soap film
r=0                    #normal incidence
t=5*10**-7             #thickness of film

#Calculations:

#Now, condition for maxima is
#2ut*cos r=(2n-1)lamda/2
lamda1=4*u*t*math.cos(r)/(2*1-1)          #n=1
print"For n=1 wavelength is =",lamda1,"m"
lamda2=4*u*t*math.cos(r)/(2*2-1)          #n=2
print"For n=2 wavelength is =",lamda2,"m"
lamda3=4*u*t*math.cos(r)/(2*3-1)          #n=3
print"For n=3 wavelength is =",lamda3,"m"
lamda4=4*u*t*math.cos(r)/(2*4-1)          #n=4
print"For n=4 wavelength is =",lamda4,"m"

print"Out of these wavelengths wavelength for n=3 lies in the visible spectrum."
print"Hence, wavelength for n=3 is the most reflected wavelength."

For n=1 wavelength is = 2.66e-06 m
For n=2 wavelength is = 8.86666666667e-07 m
For n=3 wavelength is = 5.32e-07 m
For n=4 wavelength is = 3.8e-07 m
Out of these wavelengths wavelength for n=3 lies in the visible spectrum.
Hence, wavelength for n=3 is the most reflected wavelength.

Example 1.22, Page number 1-46

In [33]:
import math

#Given Data:
u=1.5                #Refractive index of a oil
lamda=5.88*10**-7    #wavelength of required yellow light
n=1                  #for smallest thickness
r=60*math.pi/180     #angle of reflection

#Calculations:

#Now, condition for dark fringe is
#2ut*cos r=n*lamda
t=n*lamda/(2*u*math.cos(r))      #thickness of film
print"Thickness of the film is =",t,"m"

Thickness of the film is = 3.92e-07 m

Example 1.23, Page number 1-46

In [35]:
import math

#Given Data:
theta=20./3600*math.pi/180     #angle of wedge in radians
B=0.25*10**-2                 #fringe spacing
u=1.4                         #Refractive index of film

#Calculations:
#We know, B=lamda/(2*u*theta).
lamda=2*B*theta*u             #wavelength of light
print"Wavelength of light is =",lamda,"m"

Wavelength of light is = 6.78739153553e-07 m

Example 1.24, Page number 1-47

In [36]:
import math

#Given Data:
#Dn=2*D40

#Calculations:
#As Dn^2 = 4*n*R*lamda/u  and   Dn^2 = 4*D40^2
#i.e. 4*n*R*lamda/u = 4*4*40*R*lamda/u .hence,
n=4*40          #order of the required ring
print"Order of the dark ring which will have double the diameter of that of 40th ring is =",n
 

Order of the dark ring which will have double the diameter of that of 40th ring is = 160

Example 1.25, Page number 1-47

In [38]:
import math

#Given Data:
lamda1=6*10**-7       #wavelength of light
lamda2=4.5*10**-7    #wavelength of light
R=0.9                 #Radius of curvature

#Calculations:
#As Dn^2=4*n*R*lamda.
#Dn^2=D(n+1)^2 for different wavelengths.we get,
n=lamda2/(lamda1-lamda2)       #nth dark ring due to lam1 which coincides with (n+1)th dark ring due lamda2
D3=math.sqrt(4*n*R*lamda1)     #diameter of 3rd dark ring for lamda1
print"Diameter of 3rd dark ring for lam1 is =",D3,"m"

Diameter of 3rd dark ring for lam1 is = 0.00254558441227 m

Example 1.26, Page number 1-48

In [39]:
import math

#Given Data:
i=45*math.pi/180        #angle of incidence
u=4./3                  #Refractive index of soap film
lamda=5*10**-7          #wavelength of light
t=1.5*10**-6            #thickness of film

#Calculations:
#u=sin i/sin r        #Snell's law   .So,
r=math.asin(math.sin(i)/u)     #angle of reflection

#Now, condition for dark band is
#2ut*cos r=n*lamda
n=2*u*t*math.cos(r)/lamda    #order of band
print"order of dark band is =",n

order of dark band is = 6.78232998313

Example 1.27, Page number 1-49

In [40]:
import math

#Given Data:
D5=0.336*10**-2         #Diameter of 5th ring
D15=0.59*10**-2         #Diameter of 15th ring
lamda=5.89*10**-7       #wavelength of light
p=10                    #n=5,n+p=15

#Calculations:
#(Dn+p)^2-Dn^2=4*p*lamda*R/u
R=((D15)**2-D5**2)/(4*p*lamda)    #Radius of curvature of the lens
print"Radius of curvature of the lens is =",R,"m"

Radius of curvature of the lens is = 0.998319185059 m

Example 1.29, Page number 1-50

In [41]:
import math

#As Dn^2=4*n*R*lamda.
#thus, Dn is directly proportional to sqaure root of n
D5=math.sqrt(5)     #D5 is directly proportional to sqaure root of 5
D4=math.sqrt(4)     #D4 is directly proportional to sqaure root of 4
k1=D5-D4
print"Separation between D5 and D4 is directly proportional to =",k1

D80=math.sqrt(80)     #D80 is directly proportional to sqaure root of 80
D79=math.sqrt(79)     #D79 is directly proportional to sqaure root of 79
k2=D80-D79
print"Separation between D80 and D79 is directly proportional to =",k2

print"Thus, (D80-D79) < (D5-D4). Hence proved."

Separation between D5 and D4 is directly proportional to = 0.2360679775
Separation between D80 and D79 is directly proportional to = 0.0560774926836
Thus, (D80-D79) < (D5-D4). Hence proved.

Example 1.30, Page number 1-51

In [42]:
import math

#Given Data:
D5=0.336*10**-2       #Diameter of 5th ring
D15=0.59*10**-2       #Diameter of 15th ring
p=10                  #n=5,n+p=15
R=1                   #Radius of curvature

#Calculations:
#(Dn+p)^2-Dn^2=4*p*lamda*R/u
lamda=((D15)**2-D5**2)/(4*p*R)    #Wavelength of light
print"Wavelength of light is =",lamda,"m"

Wavelength of light is = 5.8801e-07 m

Example 1.31, Page number 1-51

In [56]:
import math


#Condition for bright band is
#2ut*cos r = (2n-1)*lamda1

#for consecutive bands, 2n=(lamda1+lamda2)/(lamda1-lamda).
#thus, 2ut*cos r = lamda2*lamda1/(lamda1-lamda2)

#And, thicknessof film 
#t= lamda2*lamda1/((2*u*cosr)(lamda1-lamda2))
print"Hence expression for thickness of film is obtained."

Hence expression for thickness of film is obtained.
In [ ]: