# CHAPTER17 : LIGHT WAVES¶

## Example E1 : Pg 361¶

In :
#example 17.1
#calculation of the speed of light in glass

#given data
mu=1.5#refractive index of glass
v0=3.*10.**8.#speed(in m/s) of light in vacuum

#calculation
v=v0/mu#definition of the refractive index

print '%s %.2f %s' %('the speed of light in glass is',v,'m/s')

the speed of light in glass is 200000000.00 m/s


## Example E2 : Pg 366¶

In :
#example 17.2
#calculation of the separation between successive bright fringes

#given data
d=0.10*10.**-3.#separation(in m) between the slits
lambd=600.*10.**-9.#wavelength(in m) of the light used
D=1.#separation(in m) between the slits and the screen

#calculation
w=D*lambd/d#separation between successive bright fringes

print '%s %3.1e %s %.2f %s' %('the separation between successive bright fringes is',w,'m or',w*10**3,'mm')

the separation between successive bright fringes is 6.0e-03 m or 6.00 mm


## Example E3 : Pg 367¶

In :
#example 17.3
#calculation of the wavelength of light in the water

#given data
lambdan=589.#wavelength(in nm) of light in vacuum
mu=1.33#refractive index of water

#calculation
lambd=lambdan/mu#definition of the refractive index

print '%s %.2f %s' %('the wavelength of light in the water is',round(lambd),'nm')

the wavelength of light in the water is 443.00 nm


## Example E4 : Pg 369¶

In :
#example 17.4
#calculation of the minimum thickness of the film

#given data
lambd=589.#wavelength(in nm) of the light used
mu=1.25#refractive index of the material

#calculation
#for strong reflection......2*mu*d = lambda/2
d=lambd/(4.*mu)#minimum thickness

print '%s %.2f %s' %('the minimum thickness of the film is',round(d),'nm')

the minimum thickness of the film is 118.00 nm


## Example E5 : Pg 372¶

In :
#example 17.5
#calculation of the angular divergence for most of the light getting diffracted

#given data
lambd=450.*10.**-9.#wavelength(in m) of the light used
b=0.2*10.**-3.#width(in m) of the slit

#calculation
#for theta tends to zero......sin(theta) = theta
theta1=lambd/b#angle of minima
theta2=-lambd/b#angle of minima
theta=theta1-theta2#angular divergence

print '%s %3.1e %s' %('the angular divergence for most of the light getting diffracted is',theta,'radian')

the angular divergence for most of the light getting diffracted is 4.5e-03 radian


## Example E6 : Pg 373¶

In :
#example 17.6
#calculation of the diameter of the disc image

#given data
lambd=590.*10.**-9.#wavelength(in m) of the light used
b=10.*10.**-2.#diameter(in m) of the converging lens used
d=20.#distance(in m) between the lens and the point of focus

#calculation
d=2.*r#diameter of the disc image

print '%s %3.1e %s' %('the diameter of the disc image is',d,'cm')

the diameter of the disc image is 2.9e-04 cm


# WORKED EXAMPLES¶

## Example E1w : Pg 376¶

In :
#example 17.1w
#calculation of the limits of wavelengths in the water

#given data
lambda01=400.#mimimum wavelength(in nm) of the light used
lambda02=700.#maximum wavelength(in nm) of the light used
mu=1.33#refractive index of water

#calculation
lambda1=lambda01/mu#definition of the refractive index
lambda2=lambda02/mu#definition of the refractive index

print '%s %.2f %s %.2f %s' %('the limits of wavelengths in the water are',lambda1,'nm and',lambda2,'nm')

the limits of wavelengths in the water are 300.75 nm and 526.32 nm


## Example E2w : Pg 376¶

In :
#example 17.2w
#calculation of the refractive index of the glass

#given data
x1=2.#distance(in cm)travelled through the glass
x2=2.25#distance(in cm)travelled through the water
muw=1.33#refractive index of water

#calculation
#for 'x' distance travelled through a medium of refractive index 'mu',the optical path is 'mu*x'
mug=muw*x2/x1#refractive index of glass

print '%s %.2f' %('the refractive index of the glass is',mug)

the refractive index of the glass is 1.50


## Example E3w : Pg 377¶

In :
#example 17.3w
#calculation of the wavelengths of the violet and the red light

#given data
D=2.5#separation(in m) between the slit and the screen
d=0.5*10.**-3.#separation(in m) between the slits
yv=2.*10.**-3.#distance(in m) between the central white fringe and the first violet fringe
yr=3.5*10.**-3.#distance(in m) between the central white fringe and the first red fringe

#calculation
lambdav=yv*d/D#wavelength of the violet light
lambdar=yr*d/D#wavelength of the red light

print '%s %.2f %s' %('the wavelength of the violet light is',lambdav*10**9,'nm')
print '%s %.2f %s' %('\nthe wavelength of the red light is',lambdar*10**9,'nm')

the wavelength of the violet light is 400.00 nm

the wavelength of the red light is 700.00 nm


## Example E4w : Pg 377¶

In :
#example 17.4w
#calculation of the separation between the slits

#given data
lambd=589.3*10.**-9.#wavelength(in m) of the sodium light
D=100.*10.**-2.#separation(in m) between the slit and the screen
n=10.#number of the bright fringe
x=12.*10.**-3.#distance(in m) between the central maximum and the tenth bright fringe

#calculation
d=n*lambd*D/x#separation between the slits

print '%s %3.1e %s %.2f %s' %('the separation between the slits is',d,'m or',d*10**3,'mm')

the separation between the slits is 4.9e-04 m or 0.49 mm


## Example E5w : Pg 377¶

In :
#example 17.5w
#calculation of the ratio of maximum intensity to the minimum intensity in the interference fringe pattern
#given data
#intensity of the light coming from one slit in Young's double slit experiment is double the intensity of the light coming from the other slit
n=2.
import math
#calculation
r=((math.sqrt(n)+1.)**2.)/((math.sqrt(n)-1.)**2.)#required ratio

print '%s %.2f' %('the ratio of maximum intensity to the minimum intensity in the interference fringe pattern is',round(r))

the ratio of maximum intensity to the minimum intensity in the interference fringe pattern is 34.00


## Example E6w : Pg 377¶

In :
#example 17.6w
#calculation of the ratio of maximum intensity to the minimum intensity in the interference pattern

#given data
#width of one slit in Young's double slit experiment is double that of the other
n=2.

#calculation
r=((n+1.)**2.)/((n-1.)**2.)#required ratio

print '%s %.2f' %('the ratio of maximum intensity to the minimum intensity in the interference pattern is',r)

the ratio of maximum intensity to the minimum intensity in the interference pattern is 9.00


## Example E7w : Pg 377¶

In :
#example 17.7w
#calculation of the maximum and the minimum path difference at the detector

#given data
lambd=600.*10.**-9.#wavelength(in m) of the light
d=1.*10.**-2.*10.**-2.#distance(in m) between the sources

#calculation
pdmax=d#path diffrence maximum
pdmin=0#path diffrence minimum
#farthest minima occurs for path difference lambda/2
#sqrt(D**2 + d**2) - D = lambda/2
D=(d**2./lambd)-(lambd/4.)#distance of the farthest minima

print '%s %3.1e %s' %('the maximum path difference on moving the detector along S1P line is',pdmax*10**2,'m')
print '%s %.2f %s' %('\nthe minimum path difference on moving the detector along S1P line is',pdmin*10**2,'cm')
print '%s %.2f %s' %('\nthe farthest minimum is located at a distance of',D*10**2,'cm from the point S1')

the maximum path difference on moving the detector along S1P line is 1.0e-02 m

the minimum path difference on moving the detector along S1P line is 0.00 cm

the farthest minimum is located at a distance of 1.67 cm from the point S1


## Example E8w : Pg 378¶

In :
#example 17.8w
#calculation of the distance of bright fringe from the central maximum

#given data
lambda1=6500.*10.**-10.#wavelength(in m) of the light beam1
lambda2=5200.*10.**-10.#wavelength(in m) of the light beam2
d=2.0*10.**-3.#separation(in m) between the slits
D=120.*10.**-2.#separation(in m) between the slits and the screen
n=3.#number of the bright fringe

#calculation
y=n*lambda1*D/d#the distance of bright fringe from the central maximum
#from the equation of the distance of bright fringe from the central maximum.....y=n*lambda*D/d
#let m th bright fringe of beam 1 coincides with n th bright fringe of beam 2
#ym = yn
#m : n = 4 : 5.....is their minimum integral ratio
m=4.
ym=m*lambda1*D/d#least distance from the central maximum where both wavelengths coincides
print '%s %.2f %s' %('the distance of the third bright fringe from the central maximum is',y*10**2,'cm')
print '%s %.2f %s' %('\nthe least distance from the central maximum where both the wavelengths coincides is',ym*10**2,'cm')

the distance of the third bright fringe from the central maximum is 0.12 cm

the least distance from the central maximum where both the wavelengths coincides is 0.16 cm


## Example E9w : Pg 378¶

In :
#example 17.9w
#calculation of the number of fringes that will shift due to introduction of the sheet

#given data
lambd=600.*10.**-9.#wavelength(in m) of the light used
t=1.8*10.**-5.#thickness(in m) of the transparent sheet
mu=1.6#refractive index of the material

#calculation
n=((mu-1.)*t)/lambd#number of fringes shifted

print '%s %.2f' %('the number of fringes that will shift due to introduction of the sheet is',n)

the number of fringes that will shift due to introduction of the sheet is 18.00


## Example E10w : Pg 378¶

In :
#example 17.10w
#calculation of the wavelengths in the visible region that are strongly reflected

#given data
d=.5*10.**-6.#thickness(in m) of the glass plate
mu=1.5#refractive index of the medium
lambda1=400.*10.**-9.#minimum wavelength(in m) of the visible region
lambda2=700.*10.**-9.#maximum wavelength(in m) of the visible region

#calculation
#condition for strong reflection of light of wavelength lambda is
#2*mu*d = (n + (1/2))*lambda............(1)
n1=round((2.*mu*d/lambda1)-(1./2.))#integral value of n for lambda1
n2=round((2.*mu*d/lambda2)-(1./2.))#integral value of n for lambda2
lambda1n=(2.*mu*d)/(n1+(1./2.))#from equation (1)
lambda2n=(2.*mu*d)/(n2+(1./2.))#from equation (1)

print '%s %.2f %s %.2f %s' %('the wavelengths in the visible region that are strongly reflected are',round(lambda1n*10**9),'nm and',round(lambda2n*10**9),'nm')

the wavelengths in the visible region that are strongly reflected are 429.00 nm and 600.00 nm


## Example E11w : Pg 378¶

In :
#example 17.11w
#calculation of the distance between the two first order minima

#given data
b=.40*10.**-3.#width(in m) of the slit
D=40.*10.**-2.#separation(in m) between the slit and the screen
lambd=546.*10.**-9.#wavelength(in m) of the light used

#calculation
#linear distances from the central maxima are given by..x = D*tan(theta) = D*sin(theta) = +-lambda*D/b
sep=2.*lambd*D/b#separation between the minima

print '%s %.2f %s' %('the distance between the two first order minima is',sep*10**3,'mm')

the distance between the two first order minima is 1.09 mm