# 3: Interference¶

## Example number 3.1, Page number 71¶

In :
#importing modules
from __future__ import division
import math

#Variable declaration
beta = 0.51;    #Fringe width(mm)
d = 2.2;        #Distance between the slits(mm)
D = 2;      #Distance between the slits and the screen(m)

#Calculation
beta = beta*10**-1;     #Fringe width(cm)
d = d*10**-1;    #Distance between the slits(cm)
D=D*10**2;    #Distance between the slits and the screen(cm)
lamda = beta*d/D;    #Wavelength of light(cm)
lamda = lamda*10**8;     #Wavelength of light(A)

#Result
print "The wavelength of light is",lamda, "angstrom"

The wavelength of light is 5610.0 angstrom


## Example number 3.2, Page number 71¶

In :
#importing modules
from __future__ import division
import math

#Variable declaration
lambda1 = 4250;    #First wavelength emitted by source of light(A)
lambda2 = 5050;    #Second wavelength emitted by source of light(A)
D = 1.5;    #Distance between the source and the screen(m)
d = 0.025;       #Distance between the slits(mm)
n = 3;    #Number of fringe from the centre

#Calculation
lambda1 = lambda1*10**-10;     #First wavelength emitted(m)
lambda2 = lambda2*10**-10;     #Second wavelength emitted(m)
d = d*10**-3;     #Distance between the slits(m)
x3 = n*lambda1*D/d;    #Position of third bright fringe due to lambda1(m)
x3_prime = n*lambda2*D/d;    #Position of third bright fringe due to lambda2(m)
x = x3_prime-x3;      #separation between the third bright fringe(m)
x = x*10**2;    #separation between the third bright fringe(cm)

#Result
print "The separation between the third bright fringe due to the two wavelengths is",x, "cm"

The separation between the third bright fringe due to the two wavelengths is 1.44 cm


## Example number 3.3, Page number 71¶

In :
#importing modules
import math

#Variable declaration
lamda = 5.5*10**-5;    #Wavelength emitted by source of light(cm)
n = 4;    #Number of fringes shifted
t = 3.9*10**-4;    #Thickness of the thin glass sheet(cm)

#Calculation
mew = (n*lamda/t)+1;    #Refractive index of the sheet of glass
mew = math.ceil(mew*10**4)/10**4;     #rounding off the value of v to 4 decimals

#Result
print "The refractive index of the sheet of glass is",mew

The refractive index of the sheet of glass is 1.5642


## Example number 3.4, Page number 72¶

In :
#importing modules
import math

#Variable declaration
lamda = 5893;    #Wavelength of monochromatic lihgt used(A)
n = 1;    #Number of fringe for the least thickness of the film
cosr = 1;    #for normal incidence
mew = 1.42;    #refractive index of the soap film

#Calculation
#As for constructive interference,
#2*mew*t*cos(r) = (2*n-1)*lambda/2, solving for t
t = (2*n-1)*lamda/(4*mew*cosr);    #Thickness of the film that appears bright(A)
#As for destructive interference,
#2*mu*t*cos(r) = n*lambda, solving for t
t1 = n*lamda/(2*mew*cosr);    #Thickness of the film that appears bright(A)

#Result
print "The thickness of the film that appears bright is",t, "angstrom"
print "The thickness of the film that appears dark is",t1, "angstrom"

The thickness of the film that appears bright is 1037.5 angstrom
The thickness of the film that appears dark is 2075.0 angstrom


## Example number 3.5, Page number 72¶

In :
#importing modules
import math

#Variable declaration
lamda = 5893;    #Wavelength of monochromatic lihgt used(A)
n = 10;    #Number of fringe that are found
d = 1;     #Distance of 10 fringes(cm)

#Calculation
beta = d/n;    #Fringe width(cm)
lamda = lamda*10**-8;    #Wavelength of monochromatic lihgt used(cm)
theta = lamda/(2*beta);    #Angle of the wedge(rad)
theta = theta*10**4;
theta = math.ceil(theta*10**4)/10**4;     #rounding off the value of theta to 4 decimals

#Result
print "The angle of the wedge is",theta,"*10**-4 rad"

The angle of the wedge is 2.9465 *10**-4 rad


## Example number 3.6, Page number 72¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
lamda = 5900;    #Wavelength of monochromatic lihgt used(A)
t = 0.010;    #Spacer thickness(mm)
l = 10;    #Wedge length(cm)

#Calculation
t = t*10**-1;    #Spacer thickness(cm)
theta = t/l;    #Angle of the wedge(rad)
lamda = lamda*10**-8;    #Wavelength of monochromatic lihgt used(cm)
beta = lamda/(2*theta);    #Fringe width(cm)

#Result
print "The separation between consecutive bright fringes is",beta, "cm"

The separation between consecutive bright fringes is 0.295 cm


## Example number 3.7, Page number 72¶

In :
#importing modules
import math

#Variable declaration
D4 = 0.4;    #Diameter of 4th dark ring(cm)
D12 = 0.7;    #Diameter of 12th dark ring(cm)

#Calculation
#We have (dn_plus_k**2)-Dn**2 = 4*k*R*lamda
#D12**2-D4**2 = 32*R*lamda and D20**2-D12**2 = 32*R*lamda for k = 8
#since RHS are equal, by equating the LHS we get D12**2-D4**2 = D20**2-D12**2
D20 = math.sqrt((2*D12**2)-D4**2);    #Diameter of 20th dark ring(cm)
D20 = math.ceil(D20*10**4)/10**4;     #rounding off the value of D20 to 4 decimals

#Result
print "The diameter of 20th dark ring is",D20, "cm"

The diameter of 20th dark ring is 0.9056 cm


## Example number 3.8, Page number 73¶

In :
#importing modules
import math
from __future__ import division

#Variable declaration
Dn = 0.30;    #Diameter of nth dark ring with air film(cm)
dn = 0.25;    #Diameter of nth dark ring with liquid film(cm)

#Calculation
mew = (Dn/dn)**2;    #Refractive index of the liquid

#Result
print "The refractive index of the liquid is", mew

The refractive index of the liquid is 1.44


## Example number 3.9, Page number 73¶

In :
#importing modules
import math

#Variable declaration
x = 0.002945;    #Distance through which movable mirror is shifted(cm)
N = 100;    #Number of fringes shifted

#Calculation
x = x*10**-2;    #Distance through which movable mirror is shifted(m)
lamda = 2*x/N;   #Wavelength of light(m)
lamda = lamda*10**10;    #Wavelength of light(A)

#Result
print "The wavelength of light is",lamda, "angstrom"

The wavelength of light is 5890.0 angstrom


## Example number 3.10, Page number 73¶

In :
#importing modules
import math

#Variable declaration
lambda1 = 5896;    #Wavelength of D1 line of sodium(A)
lambda2 = 5890;    #Wavelength of D2 line of sodium(A)

#Calculation
lamda = (lambda1+lambda2)/2;
x = (lamda**2)/(2*(lambda1-lambda2));    #Shift in movable mirror of Michelson Interferometer(A)
x = x*10**-7;           #Shift in movable mirror of Michelson Interferometer(mm)
x = math.ceil(x*10**4)/10**4;     #rounding off the value of D20 to 4 decimals

#Result
print "The shift in movable mirror is",x, "mm"

The shift in movable mirror is 0.2894 mm

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