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"
```

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"
```

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"
```

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"
```

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
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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"
```

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
```

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"
```

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
```

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"
```

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"
```

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"
```

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"
```

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"
```

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."
```

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"
```

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"
```

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
```

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"
```

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
```

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"
```

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."
```

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"
```

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."
```

In [ ]:

```
```