# Chapter 5: Physical Optics¶

## Example 5.1, Page 297¶

In [2]:
#Variable Declaration
n1 = 10;  # Order of interference maximum for lambda = 7000 angstrom
lambda1 = 7000;  # Wavelength of the light, angstrom
lambda2 = 5000;  # Wavelength of the light, angstrom

#Calculations
# As W = D*lambda/(2*d) then, x = n1*D*lambda1/(2*d) = n2*D*lambda2/(2*d), solving for n2
n2 = n1*lambda1/lambda2;    # Order of interference maximum for lambda = 5000 angstrom

#Result
print "The order of interference maximum for wavelength of 5000 angstrom = %2d "%n2

The order of interference maximum for wavelength of 5000 angstrom = 14


## Example 5.2, Page 297¶

In [4]:
from math import *

#Variable Declaration
D = 1.6;  # Distance between the slit and the screen, m
a = 0.4;  # Distance between the slit and the biprism, m
mu = 1.52;  # Refractive index of the material of biprism
W = 1e-004;    # Fringe width, m
lamda = 5.893e-007;  # Wavelength of light used, m

#Calculations
# As W = lambda*D/(2*a(mu-1)*alpha then
alpha = ((lamda*D)/(2*a*(mu-1)*W))*180/pi;    # Angle of biprism, degrees

#Result
print "The angle of the biprism = %3.1f degrees"%alpha

The angle of the biprism = 1.3 degrees


## Example 5.3, Page 298¶

In [54]:
#Variable Declaration
lamda = 5.890e-7;  # Wavelength of source of light, m
mu = 1.6;  #refractive index of the mica sheet

#Calculations
# As del_x = W*(mu-1)*t/lambda, where del_x = 3*W, solving for t
t = 3*lamda/(mu-1);    # Thickness of the mica sheet, m

#Result
print "The thickness of the mica sheet = %5.3e cm"%(t/1e-02)


The thickness of the mica sheet = 2.945e-04 cm


## Example 5.4, Page 298¶

In [7]:
#Variable Declaration
lamda = 6.0e-7;  # Wavelength of the monochromatic light, m
D = 1;  # Distance between the screen and the two coherent sources, m
W = 5e-004;  # Fringe width, m

#Calculations
d = lamda*D/(W*1e-03);    # Distance between two coherent sources, mm

print "The distance between the two coherent sources = %3.1f mm"%d

The distance between the two coherent sources = 1.2 mm


## Example 5.5, Page 298¶

In [59]:
from math import *

#Variable Declaration
D = 1;  # Distance between slits and the screen, m
mu = 1.5;    # Refractive index of the material of biprism
a = 0.5;  # The distance between the slit and the biprism, m
W = 1.35e-004;  # Width of the fringes, m

#Calculations
alpha = (180.-179.)/2*pi/180;    # Acute angle of biprism, radian
lamda = 2*a*(mu-1)*alpha*W/D;    # Wavelength of light used, m

#Result
print "The wavelength of light used = %4d angstrom"%(lamda/1e-10)

The wavelength of light used = 5890 angstrom


## Example 5.6, Page 299¶

In [11]:
#Variable Declaration
lamda = 6.328e-007;  # Wavelength of the monochromatic light, m
D = 40;  # Distance between the slits and the screen, m
W = 0.1;  # Distance between the interference maxima, m

#Calculations
d = lamda*D/W;    # Distance between the slits, m

#Result
print "The distance between the slits = %6.4f mm"%(d/1e-03)

The distance between the slits = 0.2531 mm


## Example 5.7, Page 299¶

In [60]:
#Variable Declaration
lamda = 5.0e-007;  # Wavelength of the monochromatic light, m
D = 1;  # Distance between the silts and the screen, m
d = 5e-004/2;  # Half of the distance between the two slits, m
mu = 1.5;  # Refractive index of glass
t = 1.5e-006;  # Thickness of thin glass plate, m

#Calculations
del_x = D*(mu-1)*t/(2*d);

#Result
print "The lateral shift of central maximum = %3.1f mm"%(del_x/1e-03)

The lateral shift of central maximum = 1.5 mm


## Example 5.8, Page 300¶

In [13]:
from math import *

#Variable Declaration
lamda = 6.0e-007;  # Wavelength of the light, m
mu = 1.463;  # Refrctive index of a soap bubble film
n = 0;    # Value of n for smallest thickness
r = 0;    # Angle of refraction for normal incidence

#Calculations
# As 2*mu*t*cos(r) = (2*n+1)*lambda/2, solving for t
t = (2*n+1)*lamda/(4*mu*cos(r));    # The thickness of a soap bubble film, m

#Result
print "The thickness of a soap bubble film = %5.1f angstrom"%(t/1e-010)

The thickness of a soap bubble film = 1025.3 angstrom


## Example 5.9, Page 300¶

In [14]:
#Variable Declaration
D5 = 3.36e-003;  # Diameter of Newton's 5th ring, m
D15 = 5.90e-003;  # Diameter of Newton's 15th ring, m
m = 10;    # Number of ring
R = 1;  # Radius of the plano-convex lens, m

#Calculations
lamda = (D15**2-D5**2)/(4*m*R);

print "The wavelength of the light used = %4d angstrom"%(lamda/1e-010)

The wavelength of the light used = 5880 angstrom


## Example 5.10, Page 301¶

In [15]:
#Variable declaration
D10 = 0.005;  # Diameter of Newton's 5th ring, m
n = 10;  # Order of the ring
lamda = 6.0e-007;  # Wavelength of the light used, m

#Calculations&Results
R = (D10**2)/(4*n*lamda);    # Radius of the curvature of the lens, m
print "The radius of the curvature of the lens = %6.4f m"%R
t = D10**2/(8*R);
print "The thickness of the corresponding air film = %3.1e m"%t

The radius of the curvature of the lens = 1.0417 m
The thickness of the corresponding air film = 3.0e-06 m


## Example 5.11, Page 301¶

In [64]:
from math import *

#Variable declaration
mu = 1.43;  # Refractive index of the soap film
n = 0;    # Order of fringes for smallest thickness
i = 30;    # Angle of incidence, degrees

#Calculations
# As sin(i)/sin(r) = mu, cos(r)
cosr = sqrt(1-(sin(i*pi/180)/mu)**2);    # Cosine of angle r
lamda = 6.0e-007;  # Wavelength of the light, m
t = (2*n+1)*lamda/(4*mu*cosr);  # Thickness of the soap film, m

#Result
print "The thickness of the soap film = %4.2e m"%t

The thickness of the soap film = 1.12e-07 m


## Example 5.12, Page 301¶

In [66]:
from math import *

#Variable declaration
lamda = 5.893e-007;  # Wavelength of the sodium light, m
mu = 1.42;  # Refractive index of the soap film
r = 0;    # Angle of refraction, degrees
n = 0;    # Order of diffraction for least thickness of dark film

#Calculations&Result
t = (2*n+1)*lamda/(4*mu*cos(r));    # Least thickness of the film that will apear bright, m
print "The least thickness of the film that will appear bright = %5.1f m"%(t/1e-010)
n = 1;    # Order of diffraction for least thickness of bright film
t = n*lamda/(2*mu*cos(r));    # Least thickness of the film that will apear dark, m
print "The least thickness of the film that will appear dark = %6.2f m"%(t/1e-010)  #incorrect answer in the textbook

The least thickness of the film that will appear bright = 1037.5 m
The least thickness of the film that will appear dark = 2075.00 m


## Example 5.13, Page 302¶

In [19]:
#Variable declaration
lamda = 5.893e-007;  # Wavelength of the sodium light, m

#Calculations
# As fringe width of the thin wedge-shaped air film is
# W = lambda/(2*t/20*W), solving for t
t = (10*lamda);    # Thickness of the wire separating edges of two plane glass surfaces, m

#Result
print "The thickness of the wire = %5.3e m"%t

The thickness of the wire = 5.893e-06 m


## Example 5.14, Page 303¶

In [20]:
#Variable declaration
lamda = 5.9e-007;  # Wavelength of the reflected light, m
n = 10;  # Order of the ring
D10 = 0.005;  # Diameter of the 10th ring,in m

#Calculations&Result
R = (D10**2)/(4*n*lamda);    # Radius of curvature of the lens, m
print "The radius of curvature of the lens = %6.4f m"%R
t = (D10**2)/(8*R);    # Thickness of the corresponding air film, m
print "The thickness of the corresponding air film = %4.2e m"%t

The radius of curvature of the lens = 1.0593 m
The thickness of the corresponding air film = 2.95e-06 m


## Example 5.16, Page 304¶

In [69]:
from math import *

#Variable declaration
lamda = 6.328e-007;  # Wavelength of monochromatic light from He laser, m
n1 = 1;  # First order
n2 = 2;  # Second order
l = 6000;  # Lines/cm of the diffraction grating
A= 1.66e-6;

#Calculations&Result
theta = degrees(asin(n1*lamda/A));
print "The first order maximum angle = %4.1f degrees"%theta
theta = degrees(asin(n2*lamda/A));
print "The second order maximum angle = %4.1f degrees"%theta

The first order maximum angle = 22.4 degrees
The second order maximum angle = 49.7 degrees


## Example 5.17, Page 305¶

In [22]:
from math import *

#Variable declaration
a = 1;    # For simplicity assume slit width to be unity, unit
theta = 1; # For simplicity assume diffraction angle to be unity, unit

#Calculations
# As a*sin(theta) = m*lambda, solving for lambdas
lambda1 = a*sin(theta);    # First wavelength, angstrom
lambda2 = a*sin(theta)/2;    # First wavelength, angstrom

#Result
print "lambda1 = %d*lambda2"%(lambda1/lambda2)

lambda1 = 2*lambda2


## Example 5.18, Page 305¶

In [85]:
from math import *

#Variable declaration
lamda = 5.5e-7;  # Wavelength of light, m
a = 2.2e-6;  # Width of the slit, m
l = 6000;  # Lines /cm of the diffraction grating
# In a single slit diffraction pattern the directions of minimum intensity are given by a*sintheta = m*lambda where m = 1,2,3
# For m = 1

#Calculations&Results
m = 1;  # First order
theta = degrees(asin((m*lamda)/a));    # Angular position of first minima on either side of the central maxima, degrees
print "The angular position of first minima on either side of the central maxima = %.2f degrees"%(theta)

# For m = 2
m = 2;  # Second order
theta = degrees(asin(m*lamda/a));
print "The angular position of second minima on either side of the central maxima = %.2f degrees"%(theta)

The angular position of first minima on either side of the central maxima = 14.48 degrees
The angular position of second minima on either side of the central maxima = 30.00 degrees


## Example 5.19, Page 306¶

In [94]:
#Variable declaration
D = 1.7;  # Distance between the slit and the screen, m
W = 2.5e-003;  # Given fringe width, m
a = 8e-005;  # Width of the first slit, m
b = 4e-004;  # Width of the second slit, m
n = b;    #
p = [1, 2, 3, 4, 5, 6];

#Calculations&Results
# In a double slit experiment Fraunhoffer diffraction pattern,the fringe width is given by W = lambda*D/n
lamda = b*W/D;    # Wavelength of the light used, m
print "The wavelength of light = %4d angstrom"%(lamda/1e-010)
print "The missing orders are:\n"
for i in range(1,6):
s = ((a+b)/a)*i;
print "%d,"%s,


The wavelength of light = 5882 angstrom
The missing orders are:

6, 12, 18, 24, 30,


## Example 5.20, Page 306¶

In [95]:
#Variable declaration
D = 2;  # Distance of the screen from the slit, m
x = 1.6e-02;  # Position of centre of the second dark band, m
m = 2;    # Order of diffraction
a = 1.4e-04;  # Width of the slit, m

#Calculations
lamda = (a*x)/(m*D);    # Wavelength of light, m

#Result
print "The wavelength of the light = %4d angstrom"%((lamda/1e-010))
#rounding-off error

The wavelength of the light = 5599 angstrom


## Example 5.21, Page 307¶

In [30]:
#Variable declaration
lambda1 = 5890;  # Wavelength of the line, angstrom
lambda2 = 5896;  # Wavelength of the line, angstrom

#Calculations
d_lambda = lambda2 - lambda1;  # Wavelength difference, angstrom
n = 2;  # Order of diffraction
N = lambda2/(n*d_lambda);    # Minimum no. of lines in a grating

#Result
print "The minimum number of lines in the grating = %3d lines"%N

The minimum number of lines in the grating = 491 lines


## Example 5.22, Page 307¶

In [96]:
from math import *

#Variable declaration
lamda = 5.0e-07;  # Wavelength of the radiation, m
a_plus_b = 2.54e-02/2620;  # The grating element, m
theta_max = 90;    # Maximum value of angle of diffraction, degrees

#Calculations
n_max = a_plus_b/lamda*sin(theta_max*pi/180);    # Maximum number of visible orders

#Result
print "The number of visible orders = %2d "%n_max

The number of visible orders = 19


## Example 5.23, Page 307¶

In [11]:
from math import *

#Variable declaration
lambda1 = 6000;  # Wavelength of yellow line, angstrom
lambda2 = 4800;  # Wavelength of blue line, angstrom

#Calculations
#(a+b)sin(theta) = n*6000   ---1
#(a+b)sin(theta) = (n+1)*4800
#Comparing 1 and 2, we get the following,
n = 48./12
theta = 3./4;    # Angle of diffraction, radian
a_plus_b = (n*lambda1)/theta

#Results
print "Grating element = %d A"%a_plus_b

Grating element = 32000 A


## Example 5.26, Page 310¶

In [33]:
from numpy import *

#Variable declaration
n = 5;  # Order for given wavelength
m = [4, 5, 6, 7, 8];    # Orders of spectral lines in the visible range
lambda1 = 6000;  # Wavelength of the spectral line in visible range, angstrom
lambda2 = zeros(5);

print "The spectral lines in visible ranges are:\n"
for i in range(1,5):
l2 = (n*lambda1)/m[i];
lambda2[i] = l2;    # Preserve the lambda value
print "%4d angstrom\n"%(l2),

print "The other spectral lines in the visible range 4000A to 7000A are"
for i in range(1,5):
if lambda2[i] < 7000 and lambda2[i] > 4000:
if lambda2[i] == 6000:
continue
#print "%4dA"%lambda2[i]

The spectral lines in visible ranges are:

6000 angstrom
5000 angstrom
4285 angstrom
3750 angstrom
The other spectral lines in the visible range 4000A to 7000A are


## Example 5.27, Page 310¶

In [35]:
#Variable declaration
N = 4500;  # Number of lines in grating
n = 2;  # Order of diffraction
lambda1 = 5890;  # Wavelength, angstrom
lambda2 = 5896;  # Wavelength, angstrom

#Calculations&Result
RP2 = n*N;    # Resolving power of grating in the second order
lamda = (lambda1+lambda2)/2;    # Mean wavelength of sodium light, angstrom
d_lambda = lambda2 - lambda1;    # Wavelength difference, angstrom
RP = lamda/d_lambda;    # Calculated resolving power of grating
if RP2 <> RP:
print("The D1 and D2 lines of Na light cannot be resolved in second order");

The D1 and D2 lines of Na light cannot be resolved in second order


## Example 5.28, Page 311¶

In [37]:
#Variable declaration
lamda = 5.5e-07;  # Wavelength of light used, m
f = 3.0;  # Focal length of telescope objective, m
a = 0.01;  # Diameter of the telescope objective, m

#Calculations
# As x/f = 1.22*lambda/a, the Rayleigh criterian for resolution, solving for x
x = 1.22*f*lamda/a;    # Distance between two stars just seen as separate, m

print "The distance between two stars just seen as separate = %3.1e m "%x

The distance between two stars just seen as separate = 2.0e-04 m


## Example 5.29, Page 311¶

In [38]:
#Variable declaration
lamda = 5.461e-07;  # Wavelength of light used, m
d = 4.0e-07;  # Distance between the two luminous objects, m

#Calculations
# As d = 1.22*lambda/(2*mu*sin(alpha)) = 1.22*lambda/(2*NA), solving for NA
NA = 1.22*lamda/(2*d);     # Numerical aperature of the objective of microscope

print "The numerical aperature of the objective of microscope = %5.3f "%NA

The numerical aperature of the objective of microscope = 0.833


## Example 5.30, Page 312¶

In [39]:
#Variable declaration
lamda = 6.0e-07;  # Wavelength of light used, m
d_theta = 2.44e-06;  # Angular separation between the two stars, radian

#Calculations
a = 1.22*lamda/d_theta;    # Aperature of the objective of a telescope from Rayleigh criterian, m

print "The aperature of the objective of the telescope = %3.1f m "%a

The aperature of the objective of the telescope = 0.3 m


## Example 5.31, Page 312¶

In [40]:
#Variable declaration
lamda = 5.5e-007;  # Wavelength of light used, m
x = 1.5e-003;  # Distance between the two pinholes, m
a = 4.0e-003;  # Diameter of objective, m

#Calculations
D = a*x/(1.22*lamda);    # Minimum distance from the telescope at which the the pinhole can be resolved from Rayleigh criterian, m

print "The minimum distance from the telescope at which the the pinhole can be resolved = %4.2f m "%D

The minimum distance from the telescope at which the the pinhole can be resolved = 8.94 m


## Example 5.32, Page 312¶

In [41]:
#Variable declaration
lamda = 5.461e-07;  # Wavelength of light used, m
d = 5.55e-07;  # Distance between the two luminous objects, m

#Calculations
# As d = 1.22*lambda/(2*mu*sin(alpha)) = 1.22*lambda/(2*NA), solving for NA
NA = 1.22*lamda/(2*d);     # Numerical aperature of the objective of microscope

print "The numerical aperature of the objective of microscope = %4.2f "%NA

The numerical aperature of the objective of microscope = 0.60


## Example 5.33, Page 313¶

In [31]:
from math import *

#Variable declaration
i = 60;    # Angle of incidence, degrees
mu = tan(i*pi/180);    # Brewester's Law to calculate refractive index
A = 60*pi/180;  # Angle of prism, degrees

#Calculations
# As mu = sind((A+delta_m)/2)/sind(A/2), solving for delta_m
delta_m = 2*degrees(asin(mu*sin(A/2)))   # Angle of minimum deviation for green light for its passage through a prism, degrees

#Result
print "The angle of minimum deviation for green light for its passage through a prism = %.f degrees"%((delta_m-60))

The angle of minimum deviation for green light for its passage through a prism = 60 degrees


## Example 5.34, Page 313¶

In [43]:
#Variable declaration
lamda = 5.89e-07;  # Wavelength of light used, m
mu_O = 1.55;    # Refractive index of ordinary light
mu_E = 1.54;    # Refractive index of extraordinary light

#Calculations
tQ = lamda/(4*(mu_O-mu_E));    # The thickness of the quarter wave plate, m

print "The thickness of the quarter plate is = %6.4e m"%tQ


The thickness of the quarter plate is = 1.4725e-05 m


## Example 5.35, Page 314¶

In [44]:
#Variable declaration
theta = 9.9;  # Optical rotation of solution, degrees
l = 20;  # Length of the tube, cm
S = 66;  # Specific rotation of pure sugar solution, degree per dm-(g/cc)

#Calculations
# As the specific rotation, S = 10*theta/l*c, solving for c
c = 10*theta/(l*S);    # Concentration of solution for pure sugar, g/cc
c_prime = 0.080;    # Concentration of solution for impure sugar, g/cc
Percentage_purity = c*100/c_prime;    # Percentage purity of sugar sample

print "The percentage_purity of the sugar sample = %5.2f percent"%Percentage_purity

The percentage_purity of the sugar sample = 93.75 percent


## Example 5.36, Page 314¶

In [45]:
#Variable declaration
theta = 26.4;  # Optical rotation of sugar solution, degrees
l = 20;  # Length of the tube, cm
c = 0.20;  # Concentration of the solution, g/cc

#Calculations
S = 10*theta/(l*c);    # The specific rotation of the sugar solution, degree per dm per (g/cc)

print "The specific rotation of the sugar solution = %2d degrees"%S

The specific rotation of the sugar solution = 66 degrees


## Example 5.37, Page 315¶

In [43]:
from math import *
# Function to convert degrees to deg-min
def deg_to_dms(deg):
d = int(deg)
md = abs(deg - d) * 60
md = round(md)
return [d, md]

lamda = 7.62e-07;  # Wavelength of the polarized light, m
mu_R = 1.53914;    # Refractive index of quartz for right-handed circularly polarized light
mu_L = 1.53920;    # Refractive index of quartz for left-handed circularly polarized light
t = 5.0e-004;  # Thickness of the plate, m

theta = pi*t*(mu_L-mu_R)/lamda;     # The angle of optical rotation, radian
d = deg_to_dms(theta*180/pi);    # Call the conversion function

print "The angle of rotation produced by its plate = ",d

The angle of rotation produced by its plate =  [7, 5.0]


## Example 5.38, Page 315¶

In [49]:
#Variable declaration
theta = 13.;  # Optical rotation of the solution, degrees
l = 20.;  # Length of the tube, cm
l_prime = 30.;  # New length of the tube, cm
c = 1.;    # For simplicity assume concentration of sugar solution to be unity, g/cc

#Calculations
c_prime = c/3;    # New concentration of sugar solution, g/cc
# As, S = 10*theta/(l*c) so 10*theta/(l*c) = 10*theta_prime/(l_prime*c_prime)
# Solving for theta_prime
theta_prime = theta/(l*c)*l_prime*c_prime;    # The optical rotation produced by new length of sugar solution, degrees

print "The optical rotation of %d cm length of sugar solution = %3.1f degrees"%(l_prime, theta_prime)

The optical rotation of 30 cm length of sugar solution = 6.5 degrees


## Example 5.39, Page 315¶

In [51]:
#Variable declaration
theta = 11.;  # Optical rotation of sugar solution, degrees
l = 20;  # Length of the tube, cm
S = 66;  # Specific rotation of sugar solution, degrees

#Calculations
c = theta*10/(l*S);    # The concentration of sugar solution, g/cc

print "The strength of the solution = %6.4f g/cc"%c

The strength of the solution = 0.0833 g/cc


## Example 5.40, Page 316¶

In [53]:
 #Variable declaration
theta = 20;  # Optical rotation of sugar solution, degrees
theta_prime = 35.;  # New optical rotation of sugar solution, degrees
c = 5;  # Percentage concentration of the solution
c_prime = 10;  # New percentage concentration of the solution
l = 1;       # For simplicity assume length of the sugar solution to be unity

#Calculations
l_prime = theta_prime*l*c/(c_prime*theta);

print "The length of sugar solution for %d percent concentration and %d degrees optical rotation = %5.3f*l "%(c_prime, theta_prime, l_prime)


The length of sugar solution for 10 percent concentration and 35 degrees optical rotation = 0.875*l