Introduction To Special Relativity And Space Science (By S.P. Singh)

CHAPTER NUMBER 1 : Interference Diffraction and Polarization

EXAMPLE 1.1 : (Page Number 46)

In [16]:
import math 
#Given that
Beta=0.10#fringe width in cm
D=200# separation between source and screen in cm
lambda1=0.00055# wavelength of incident light in cm 
d= (D*lambda1)/ (10*Beta)
print "   Standard formula used    beta= lambda1*D/d  "
print "  Separation between sources is ",d," cm.  "
   Standard formula used    beta= lambda1*D/d  
  Separation between sources is  0.11  cm.  

EXAMPLE 1.2 : (PAGE NUMBER 47)

In [7]:
import math
#Given that
D=80# separation between source and screen in cm
d=0.18# separation between sources in cm 
n=4# order of fringe
x_n=1.08# distance from central bright fringe in cm  
print "Standard formula used     x_n= n*lambda1*D/d "

lambda1=d*x_n/(D*n)*1e7
print  "Wavelength of light used is" ,lambda1, "Angstrom."
Standard formula used     x_n= n*lambda1*D/d 
Wavelength of light used is 6075.0 Angstrom.

EXAMPLE NUMBER 1.3 : (PAGE NUMBER 47)

In [6]:
import math
#Given that
beta=0.0320#fringe width in cm
D=100# separation between source and screen in cm
d=0.184# separation between sources in cm 
print "   Standard formula used   beta=lambda1*D/d  "
lambda1=d*beta/D*1e8
print "Wavelength of light used is" ,lambda1,"Angstrom."
   Standard formula used   beta=lambda1*D/d  
Wavelength of light used is 5888.0 Angstrom.

EXAMPLE NUMBER 1.4 : (Page Number 47)

In [8]:
import math
 #Given that
beta=0.02 #fringe width in cm
D=100 # separation between source and screen in cm
u=30 # separation between slit and convex lens in cm
I=0.7 # separation between two images of slits on screen in cm
print"   Standard formula used   beta=lambda1*D/d  " 
v=100-u
O=I*u/v
d=O
lambda1=d*beta/D*1e8
print"  Wavelength of light used is",lambda1, "Angstrom."
   Standard formula used   beta=lambda1*D/d  
  Wavelength of light used is 6000.0 Angstrom.

EXAMPLE NUMBER 1.5 : (Page Number 47)

In [2]:
import math
#Given that
x_n=1.88# fringe separation of nth fringe from central fringe in cm 
N=20# order of fringe
beta=0.02#fringe width in cm
D=120# separation between source and eyepiece in cm
d=0.076# separation between sources in cm 
print "   Standard formula used     beta= lambda1*D/d  "
beta=x_n/N # calculation of angle formed
lambda1=d*beta/D*1e8 # calculation of Wavelength of light
print "  Wavelength of light used is", round(lambda1,4) , "Angstrom."
   Standard formula used     beta= lambda1*D/d  
  Wavelength of light used is 5953.3333 Angstrom.

EXAMPLE NUMBER 1.6 : (Page Number 48)

In [4]:
import math
#Given that
mu = 1.5 #refractive index of plane glass prism
theta = pi / 180 #angle of prism
y1 = 10 #separation between slit and biprism in cm 
y2 = 100 #separation sbetween biprism and screen in cm 
lambda1 = 0.00005893#wavelength of incident light in cm 
print "  Standard formula used   Beta = (D * lambda1) / d"
d = 2 * ( mu -1) * theta * y1
D =  y1 + y2
Beta = (D * lambda1) / d
print "  Fringe width observed at distance 1 meter  is",round(Beta,4)
  Standard formula used   Beta = (D * lambda1) / d
  Fringe width observed at distance 1 meter  is 0.0371

EXAMPLE NUMBER 1.7 : (Page Number 49)

In [3]:
import math
#Given that
mu=1.52#refractive index of plane glass prism
theta1=pi/180#angle of prism
y1=25#separation between slit and biprism in cm 
y2=175#separation between biprism and screen in cm 
lambda1=0.000055#wavelength of incident light in cm 
beta1=0.02#fringe width in cm
print "   Standard formula used     beta= lambda1*D/d.  "
D=y1+y2
d= (D*lambda1)/beta1
theta1=d/(2*(mu-1)*y1)
vertex_angle=180-(2*theta1*180/pi)
print "  Vertex angle of biprism  is",round(vertex_angle,4),"degree."
   Standard formula used     beta= lambda1*D/d.  
  Vertex angle of biprism  is 177.5759 degree.

EXAMPLE NUMBER 1.8 : (Page Number 49)

In [5]:
import math
#Given that
mu=1.60#refractive index of plane glass prism
lambda1=0.0000589#wavelength of incident light in cm 
N=15#order of fringe
print "   Standard formula used     del_x = D/2d *(mu-1)*t  "
t=N*lambda1/(mu-1)
print "  Thickness of sheet is", round(t,4),"cm."
   Standard formula used     del_x = D/2d *(mu-1)*t  
  Thickness of sheet is 0.0015 cm.

EXAMPLE NUMBER 1.9 : (Page Number 50)

In [7]:
import math
#Given that
t=0.00035#thickness of glass sheet in cm
lambda1=0.000055#wavelength of incident light in cm 
N=4#order of fringe
print "   Standard formula used     (mu – 1 )*t = n* lambda1  "
mu=N*lambda1/t+1
print "  Refractive index of sheet is", round(mu,4)
   Standard formula used     (mu – 1 )*t = n* lambda1  
  Refractive index of sheet is 1.6286

EXAMPLE NUMBER 1.10 : (Page Number 50)

In [1]:
import math
#Given that
t = 5e-5 #thickness of soap film in cm 
theta1 = 35 #angle of view in degree
mu = 1.33 #refractive index of soap film 

a = 0
print "Standard formula used is  2*mu*t*cos(r) = n*lambda1 "
r = arcsin(sin(theta1 * pi /180) / mu)
for n in range(1,3):
    lambda1 = 2 * mu * t * cos(r) / n
    if lambda1 > t:
        a = a + 1
    

print "    The lowest order n =",a," will be absent in visible region."
Standard formula used is  2*mu*t*cos(r) = n*lambda1 
    The lowest order n = 2  will be absent in visible region.

EXAMPLE NUMBER 1.11 : (Page Number 50)

In [2]:
import math
#Given that
D=120#separation between source and screen in cm
d=0.00075#separation between sources in cm 
l=1.888#transverse distance moved by eyepiece in cm
N=25#order of fringe
print "   Standard formula used   beta=lambda1*D/d  "
lambda1=d*l/(D*N)*1e10
print "  Wavelength of light used is .", lambda1, "Angstrom"
   Standard formula used   beta=lambda1*D/d  
  Wavelength of light used is . 4720.0 Angstrom

EXAMPLE NUMBER 1.12 : (Page Number 51)

In [3]:
import math
#Given that
D15=0.59#diameter of 15th newton’s ring in cm
D5=0.336#diameter of 5th newton’s ring in cm
R=100#radius of Plano convex lens in cm
p=15-5
print "   Standard formula used   D_a**2 – D_b**2 = 4*p*R*lambda1  "
lambda1=(D15**2 - D5**2)/(4*p*R)*1e8
print "  Wavelength of light used is ", lambda1, "Angstrom."
   Standard formula used   D_a**2 – D_b**2 = 4*p*R*lambda1  
  Wavelength of light used is  5880.1 Angstrom.

EXAMPLE NUMBER 1.13 : (Page Number 52)

In [4]:
import math
#Given
t=40 #length of tube in cm
lambda1=5e-5 #wavelength of incident light in cm 
n=150#order of fringe
print "   Standard formula used     (mu – 1 )*t = n* lambda1  "
t=n*lambda1/t+1
print "  Refractive index of oil film is .", t
   Standard formula used     (mu – 1 )*t = n* lambda1  
  Refractive index of oil film is . 1.0001875

EXAMPLE NUMBER 1.14 : (Page Number 52)

In [5]:
import math
#Given
no_fringe = 250 #Number of fringes observed through telescope
lambda11 = 4e-5#wavelength of incident light in cm 
lambda12 = 6.5e-5 #wavelength of incident light in cm 
print "   Standard formula used    2*t = p*lambda1"
p =  no_fringe * lambda11 / (lambda12- lambda11)
t = p * lambda12 / 2
print "  Thickness  of air film  is.  ", t,"cm"
   Standard formula used    2*t = p*lambda1
  Thickness  of air film  is.   0.013 cm

EXAMPLE NUMBER 1.15 : (Page Number 52)

In [7]:
import math
#Given
mu_oil=1.3#refractive index of oil
mu_glass=1.5#refractive index of glass
lambda11=5e-7#wavelength of incident light in cm 
lambda12=7e-7#wavelength of incident light in cm 
print "   Standard formula used     2*mu*t*cos r = (p +0.5)*lambda1  "
p= ((lambda12+lambda11)/ (lambda12-lambda11))/2
t= ((p+0.5)*lambda11)/ (2*mu_oil)*1e10
x=ceil(t)
print "  Thickness of oil film  is ",round(t,4), "Angstrom."
   Standard formula used     2*mu*t*cos r = (p +0.5)*lambda1  
  Thickness of oil film  is  6730.7692 Angstrom.

EXAMPLE NUMBER 1.16 : (Page Number 53)

In [8]:
import math
#Given
lambda1=5.6e-5#wavelength of incident light in cm 
f=4#focal length in meter
mu=1.5#refractive index of glass\
n=4#order of fringe
print "   Standard formula used     D_n= sqrt(2*(2*n-1)*lambda1*R  "
R= (mu-1)*2*f
D_4=sqrt (2*(2*n-1)*lambda1*R*100)
print "Diameter of 4th bright fringe is",D_4,"cm."
   Standard formula used     D_n= sqrt(2*(2*n-1)*lambda1*R  
Diameter of 4th bright fringe is 0.56 cm.

EXAMPLE NUMBER 1.17 : (Page Number 53)

In [10]:
import math
#Given
D_5=0.336#diameter of fifth ring in cm
D_15=0.59#diameter of fifteenth ring in cm
lambda1=5.893e-5#wavelength of incident light in cm 
p=10
print "   Standard formula used     D_(n+p) **2 – D_n**2 = 4*p*R*lambda1  "
r= ((D_15**2-D_5**2)/ (4*p*lambda1))
print "   Radius of curvature of Plano-convex lens is  ",round(r,4),"cm"
   Standard formula used     D_(n+p) **2 – D_n**2 = 4*p*R*lambda1  
   Radius of curvature of Plano-convex lens is   99.7811 cm

EXAMPLE NUMBER 1.18a : (Page Number 54)

In [12]:
import math
#Given
D_10=0.5#diameter of 10th dark ring
lambda1=5.9e-5#wavelength of incident light in cm 
n=10#order of ring 
print "   Standard formula used     r_n**2 = n*lambda1*R  "
r=D_10/2
R=r**2/ (n*lambda1)/1000
print "Radius of curvature is  ",round(R,4),"m"
   Standard formula used     r_n**2 = n*lambda1*R  
Radius of curvature is   0.1059 m

EXAMPLE NUMBER 1.18b : (Page Number 54)

In [13]:
import math
#Given
lambda1=5.9e-5#wavelength of incident light in cm 
n=10#order of ring 
print "   Standard formula used     2t = n*lambda1  "
t=n*lambda1/200
print " Thickness of air film is  ",t,"m."
   Standard formula used     2t = n*lambda1  
 Thickness of air film is   2.95e-06 m.

EXAMPLE NUMBER 1.19 : (Page Number 54)

In [14]:
import math
#Given
mu=4/3
D_10=0.6#diameter of tenth ring in cm
lambda1=6.0e-5#wavelength of incident light in cm 
n=10#order of ring 
print "   Standard formula used     D_n**2 = 4*n*lambda1*R/mu  "
R= (mu*D_10**2/ (4*n*lambda1))
W=ceil(R);
print " Radius of curvature of lens is ",W,"cm." 
   Standard formula used     D_n**2 = 4*n*lambda1*R/mu  
 Radius of curvature of lens is  150.0 cm.

EXAMPLE NUMBER 1.20 : (Page Number 54)

In [15]:
import math
#Given
grating_element=6000#lines per centimeter
theta=30#angle of second order spectral line in degree 
n=2#order
print "   Standard formula used     n*lambda1= (a+b)*sin(theta)  "
lambda1=sin(theta*pi/180)/(grating_element*n)
print " Wavelength is ",lambda1,"cm."
   Standard formula used     n*lambda1= (a+b)*sin(theta)  
 Wavelength is  4.16666666667e-05 cm.

EXAMPLE NUMBER 1.21 : (Page Number 55)

In [23]:
import math
#GivenS
lambda11=6.2e-5# wavelength of monochromatic light in cm
grating_element= 0.0002# lines per centimeter(1/5000)
theta11=90# angle of second order spectral line in degree 
print "   Standard formula used     n*lambda11= (a+b)*sin(theta11)   "
n=grating_element/lambda11
print " Maximum order n = ",round(n,4),"may be seen in between the given wavelength spectrum."
   Standard formula used     n*lambda11= (a+b)*sin(theta11)   
 Maximum order n =  3.2258 may be seen in between the given wavelength spectrum.

EXAMPLE NUMBER 1.22 : (Page Number 55)

In [24]:
import math
#Given
lambda11=5.5e-5# wavelength of monochromatic light in cm
grating_element=0.00025# lines per centimeter(1/4000)
n=3# order of spectrum
print "   Standard formula used     n*lambda11= (a+b)*sin(theta11) "
sin_theta11=n*lambda11/grating_element
cos_theta11=sqrt(1-sin_theta11**2)
disp_pow=n/ (grating_element*cos_theta11)
print " Dispersive power   is    ",round(disp_pow,4)
   Standard formula used     n*lambda11= (a+b)*sin(theta11) 
 Dispersive power   is     15973.046

EXAMPLE NUMBER 1.23a : (Page Number 56)

In [32]:
import math
# Given That
lambda111=5.89e-5# wavelength in cm
lambda112=5.896e-5#wavelength in cm
n=1# for second order spectrum
t = 2 # width of detraction grating
grating_element = 425 # no. of lines per cm
print "   Standard formula used     lambda11 / d_lambda11 = n*N  "
total_line = t * grating_element
print "  Total number of lines on diffraction grating is",total_line, "So"
N=lambda111/ (lambda112-lambda111)/n
if N > total_line:
    print "   Lines will not be resolved in ",n," order ."
    print " as", round(N,4), " lines are required  for diffraction "
else: 
    print " as ",round(N,4),"  lines are required  for diffraction are.  "
    print "  Lines will be resolved in ",n," order"
   Standard formula used     lambda11 / d_lambda11 = n*N  
  Total number of lines on diffraction grating is 850 So
   Lines will not be resolved in  1  order .
 as 981.6667  lines are required  for diffraction 

EXAMPLE NUMBER 1.23b : (Page Number 56)

In [33]:
import math
# Given That
lambda111=5.89e-5# wavelength in cm
lambda112=5.896e-5#wavelength in cm
n=2# for second order spectrum
t = 2 # width of diffraction grating 
grating_element = 425 # no. of lines per cm
print "   Standard formula used     lambda11 / d_lambda11 = n*N  "
total_line = t * grating_element
print "  Total number of lines on diffraction grating",total_line,", So"
N=lambda111/ (lambda112-lambda111)/n
if N > total_line:
    print " Lines will not be resolved in",n," order."
    print " as ",round(N,4),"lines are required  for diffraction are  "
else:
    print "  Lines will be resolved in",n,"order"
    print " as ",round(N,4)," lines are required  for diffraction are . "
   Standard formula used     lambda11 / d_lambda11 = n*N  
  Total number of lines on diffraction grating 850 , So
  Lines will be resolved in 2 order
 as  490.8333  lines are required  for diffraction are . 

EXAMPLE NUMBER 1.24 : (Page Number 56)

In [35]:
import math
# Given That
lambda111=5.89e-5# wavelength in cm
lambda112=5.896e-5#wavelength in cm
t=2.5# width of grating in cm
n=2# for second order spectrum
print "   Standard formula used     lambda11 / d_lambda11 = n*N  "
N=lambda111/ (lambda112-lambda111)/n
grating_element=N/t
print " Minimum number of lines required is ",round(grating_element,4)
   Standard formula used     lambda11 / d_lambda11 = n*N  
 Minimum number of lines required is  196.3333

EXAMPLE NUMBER 1.25 : (Page Number 56)

In [36]:
import math
# Given That
a=12e-5# slit width in cm
lambda111=5.89e-5# wavelength in cm
lambda112=5.896e-5#wavelength in cm
n=2# for second order spectrum
print "   Standard formula used     lambda11 / d_lambda11 = n*N   "
d_lambda11 = lambda112-lambda111
grating_element= lambda111/ (d_lambda11*n)
print " Minimum number of lines required is ", ceil(grating_element)
   Standard formula used     lambda11 / d_lambda11 = n*N   
 Minimum number of lines required is  491.0

EXAMPLE NUMBER 1.26 : (Page Number 57)

In [37]:
import math
# Given That
a = 12e-5 # slit width in cm
lambda11 = 6e-5 # wavelength in cm
print "   Standard formula used     a*sin(theta11) = lambda11  "
theta11 = arcsin((lambda11 / a))
print " Half angular width of central bright maxima is ",ceil (theta11 * 180 / pi)," degree ."
   Standard formula used     a*sin(theta11) = lambda11  
 Half angular width of central bright maxima is  31.0  degree .

EXAMPLE NUMBER 1.27 : (Page Number 58)

In [39]:
import math
# Given That
lambda111 = 5.9e-5 # wavelength in cm
lambda112 = 5.896e-5 #wavelength in cm
lambda11 = 5.89e-5 # wavelength in cm
grating_element = 4000 # lines per cm
t = 4 # width of grating in  cm
n = 1 # for first order spectrum
print "   Standard formula used     lambda11 / d_lambda11 = n*N "
N = t * grating_element
Resolv_pow = lambda11 /(lambda112 - lambda11)
N = Resolv_pow / n
if grating_element > N: 
     print "Grating will well resolve two spectral lines.  " 
   Standard formula used     lambda11 / d_lambda11 = n*N 
Grating will well resolve two spectral lines.  

EXAMPLE NUMBER 1.28 : (Page Number 58)

In [41]:
import math
# Given That
aperture=6.4e-3# linear aperture in cm
lambda11=6.24e-5# wavelength in cm
f=50# separation between lens and screen in cm
n=1# for first order spectrum
print "   Standard formula used     a*sin(theta11 ) = lambda11  "
sin_theta11=n*lambda11/aperture
d=f*sin_theta11
print " Distance between the center and the first fringe is",ceil((d*100)/100)," cm. "
   Standard formula used     a*sin(theta11 ) = lambda11  
 Distance between the center and the first fringe is 1.0  cm. 

EXAMPLE NUMBER 1.29 : (Page Number 59)

In [42]:
import math
# Given That
theta11 = 60 # angle between plane of vibration of incident beam with optic axis
print "Standard formula used is   I = A**2*cos**2(theta11)  "
ratio = (tan(theta11*pi /180))**2    # ratio of extraordinary and aordinary intensites
print "Ratio of extraordinary and ordinary intensites is ", ratio
Standard formula used is   I = A**2*cos**2(theta11)  
Ratio of extraordinary and ordinary intensites is  3.0

EXAMPLE NUMBER 1.30 : (Page Number 59)

In [43]:
import math
# Given That
mu_e = 1.553 # refractive index of quartz plate for extra ordinary light
mu_o = 1.544 # refractive index of quartz plate for ordinary light
lambda11 = 5.89e-5 # wavelength of light in Angstrom.
print "Standard formula used is   lambda11= 2t(mu_e-mu_o)  "
t = lambda11 / (2 * (mu_e - mu_o))
print "Thickness of half wave plate of quartz is ",round(t,4)," cm."
Standard formula used is   lambda11= 2t(mu_e-mu_o)  
Thickness of half wave plate of quartz is  0.00327222222222  cm.

EXAMPLE NUMBER 1.31 : (Page Number 59)

In [45]:
import math
# Given That
lambda1=5e-5# wavelength in cm
mu_e=1.5533# refractive index for extraordinary light
mu_o=1.5422# refractive index for ordinary light
print "   Standard formula used     lambda1= 2t (mu_e-mu_o) "
t=lambda1/ (2*(mu_e-mu_o)) # calculation of Thickness of half wave plate of quartz
print "Thickness of half wave plate of quartz is",round(t,4)," cm"
   Standard formula used     lambda1= 2t (mu_e-mu_o) 
Thickness of half wave plate of quartz is 0.0023  cm

EXAMPLE NUMBER 1.32 : (Page Number 60)

In [50]:
import math
# Given That
lambda1=5.89e-5# wavelength in cm
rotation=(pi/18)# rotation of plane of polarization in degree per cm
print "   Standard formula used     delta=pi*d*del_mu/lambda1  "
del_mu=rotation*lambda1/ (pi)
print "Difference in refractive indices of substance is",round(del_mu,6)
   Standard formula used     delta=pi*d*del_mu/lambda1  
Difference in refractive indices of substance is 3e-06

EXAMPLE NUMBER 1.33 : (Page Number 60)

In [51]:
import math
# Given That
rotation=13.2# in degree
conc=0.1# gram per cubic cm
l=2# length of tube in dm
print "   Standard formula used     delta=pi*d*del_mu/lambda1  "
s= (rotation*(pi/180))/ (l*conc)
specific_rotation=s*180/pi
print "Specific rotation of sample is  ",specific_rotation,"degree."
   Standard formula used     delta=pi*d*del_mu/lambda1  
Specific rotation of sample is   66.0 degree.