2: Interference and Diffraction¶

Example number 2.1, Page number 75¶

In [2]:
#importing modules
import math
from __future__ import division

#Variable declaration
t = 12*10**-5;         #thickness of mica sheet(cm)
lamda = 6000;          #wavelength(Angstrom)
n = 1;

#Calculation
lamda = lamda*10**-10;       #wavelength(m)
mew_1 = n*lamda/t;
mew = mew_1+1;            #refractive index of mica

#Result
print "refractive index of mica is",mew
print "answer given in the book is wrong"

refractive index of mica is 1.005
answer given in the book is wrong


Example number 2.2, Page number 75¶

In [8]:
#importing modules
import math
from __future__ import division

#Variable declaration
D = 0.53;       #distance of fringes from slit(m)
lamda = 5890;      #wavelength of light(angstrom)
two_d = 0.6*10**-3;     #separation of slits(m)

#Calculation
lamda = lamda*10**-10;          #wavelength(m)
beta = D*lamda/two_d;           #width of fringes(m)
beta = beta*10**3;
beta = math.ceil(beta*10**3)/10**3;   #rounding off to 3 decimals

#Result
print "width of fringes is",beta,"*10**-3 m"

width of fringes is 0.521 *10**-3 m


Example number 2.3, Page number 75¶

In [11]:
#importing modules
import math
from __future__ import division

#Variable declaration
beta = 9*10**-4;       #width of fringes(m)
d1 = 75;          #distance of fringes from biprism(cm)
d2 = 5;           #distance of biprism from slit(cm)
lamda = 5890;      #wavelength of light(angstrom)
two_d = 0.6*10**-3;     #separation of slits(m)

#Calculation
lamda = lamda*10**-10;          #wavelength(m)
d1 = d1*10**-2;                 #distance of fringes from biprism(m)
d2 = d2*10**-2;                 #distance of biprism from slit(m)
D = d1+d2;                   #distance of fringes from slit(m)
two_d = D*lamda/beta;         #separation of slits(m)
two_d = two_d*10**4;
two_d = math.ceil(two_d*10**2)/10**2;   #rounding off to 2 decimals

#Result
print "distance between slits is",two_d,"*10**-4 m"

distance between slits is 5.24 *10**-4 m


Example number 2.4, Page number 75¶

In [13]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda = 6*10**-7;       #wavelength(m)
t = 7.2*10**-6;         #thickness(m)
n = 6;

#Calculation
mew_1 = n*lamda/t;
mew = mew_1+1;            #refractive index of sheet

#Result
print "refractive index of sheet is",mew

refractive index of sheet is 1.5


Example number 2.5, Page number 76¶

In [24]:
#importing modules
import math
from __future__ import division

#Variable declaration
beta = 3;        #fringe separation(mm)
mew = 1;         #refractive index
lamda = 6000;      #wavelength(angstrom)

#Calculation
lamda = lamda*10**-10;         #wavelength(m)
beta = beta*10**-3;         #fringe separation(m)
theta = lamda/(2*mew*beta);       #angle between plates(sec)
theeta = theta*180*3600/math.pi;     #angle between plates(sec ")
theta = theta*10**4;
theeta = math.ceil(theeta*10**3)/10**3;   #rounding off to 3 decimals

#Result
print "angle between plates is",theta,"*10**-4 sec or",theeta,"'"

angle between plates is 1.0 *10**-4 sec or 20.627 '


Example number 2.6, Page number 76¶

In [27]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda = 5900*10**-7;      #wavelength of light(m)
mew = 1;             #refractive index
n = 7.4;             #number of fringes

#Calculation
t2_t1 = n*lamda/(2*mew);       #difference of film thickness(m)
t2_t1 = t2_t1*10**2;

#Result
print "difference of film thickness is",t2_t1,"*10**-2 m"
print "answer given in the book is wrong"

difference of film thickness is 0.2183 *10**-2 m
answer given in the book is wrong


Example number 2.7, Page number 77¶

In [39]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda = 5.9*10**-7;          #wavelength of light(m)
n = 10;             #10th ring
D10 = 0.5;     #diameter of 10th ring(cm)

#Calculation
D10 = D10*10**-2;      #diameter of 10th ring(m)
R = D10**2/(4*n*lamda);        #radius of curvature of lens(m)
R = math.ceil(R*10**4)/10**4;   #rounding off to 4 decimals
t = D10**2/(8*R);           #thickness of the air film(m)

#Result
print "radius of curvature of lens is",R,"m"
print "thickness of the air film is",round(t/1e-6,2),"*10**-6 m"

radius of curvature of lens is 1.0594 m
thickness of the air film is 2.95 *10**-6 m


Example number 2.8, Page number 77¶

In [41]:
#importing modules
import math
from __future__ import division

#Variable declaration
n = 20;          #number of fringes
lamda = 5890;         #wavelength(angstrom)

#Calculation
lamda = lamda*10**-8;         #wavelength(cm)
t = n*lamda/2;                #thickness of wire(cm)
t = t*10**4;

#Result
print "thickness of wire is",t,"*10**-4 cm"

thickness of wire is 5.89 *10**-4 cm


Example number 2.9, Page number 77¶

In [46]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda = 5880;        #wavelength(angstrom)
n = 1;         #number of fringes
mew = 1.5;      #refractive index
r = 60;          #angle of refraction(degree)

#Calculation
r = r*math.pi/180;        #angle of refraction(radian)
lamda = lamda*10**-10;         #wavelength(m)
t = n*lamda/(2*mew*math.cos(r));       #smallest thickness of the plate(m)
t = t*10**10;          #smallest thickness of the plate(angstrom)

#Result
print "smallest thickness of the plate is",t,"angstrom"

smallest thickness of the plate is 3920.0 angstrom


Example number 2.10, Page number 78¶

In [50]:
#importing modules
import math
from __future__ import division

#Variable declaration
n1 = 4;          #fourth ring
n2 = 12;         #12th ring
n3 = 20;         #20th ring
D4 = 0.4;        #diameter of 4th ring(cm)
D12 = 0.7;       #diameter of 12th ring(cm)

#Calculation
p1 = n2-n1;
p2 = n3-n2;
#D12**2-D4**2 = 4*p1*lamda*R and D20**2-D12**2 = 4*p2*lamda*R
#therefore D12**2-D4**2 = D20**2-D12**2
D20 = math.sqrt((2*D12**2)-(D4**2));       #diameter of 20th ring(cm)
D20 = math.ceil(D20*100)/100;   #rounding off to 2 decimals

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

diameter of 20th ring is 0.91 cm


Example number 2.11, Page number 78¶

In [53]:
#importing modules
import math
from __future__ import division

#Variable declaration
lamda1 = 6*10**-5;        #wavelength of light 1(cm)
lamda2 = 4.5*10**-5;      #wavelength of light 2(cm)
R = 90;          #radius of curvature(cm)

#Calculation
n = lamda2/(lamda1-lamda2);           #number of fringes
Dn = math.sqrt(4*n*lamda1*R);         #diameter of nth ring(cm)
Dn = math.ceil(Dn*10**4)/10**4;   #rounding off to 4 decimals

#Result
print "diameter of nth ring is",Dn,"cm"

diameter of nth ring is 0.2546 cm