Chapter 11: WAVE OPTICS

11.1 Interference

Physical (wave) optics treats light as a wave in the study of the some light

phenomena, such as interference, diffraction and polarization.

Two waves (of the same wavelength) are said to be in phase if the crests (and

troughs) of one wave coincide with the crests (and troughs) of the other.

In this case the resultant wave would have twice the amplitude of the individual

waves - one says that constructive interference has occurred.

If the crest of one wave coincides with the trough of the second wave, they are

said to be completely out of phase.

In this case the two waves would cancel each other out - one says that

destructive interference has occurred.

Conditions for interference between two sources of light:

1. The sources must be coherent, that is they must maintain a constant

phase with respect to each other.

2. The sources must have identical wavelengths, amplitudes and

frequencies.

3. The superposition principle must apply.

To produce coherent light sources, use a single wavelength source to illuminate a

screen containing two small slits.

The light emerging from the two slits is coherent because a single source

produces the original light beam and the slits serve only to separate the original

beam into two parts.

2

11.2 Young's Double Slit Experiment Young’s double slit experiment provides evidence of the wave nature of light and

a way to measure a wavelength of light ( 10-7m).

This is a classic example of interference effects in light waves.

Two light rays pass through two slits, separated by a distance d and strike a

screen a distance, L , from the slits, as in Fig.

Figure 11.1 A schematic diagram of Young’s double-slit experiment.

The points of constructive interference will appear as bright bands on the

screen and the points of destructive interference will appear as dark bands.

Interference pattern consist of equally spaced bright fringes separated by

equally spaced dark fringes.

The condition for interference is determined by the path length difference (

L ) of the two waves or the difference of the distance travel.

If d < < L then the difference in path length r1 - r2 travelled by the two rays is

approximately:

L = r1 - r2 = d sin Path difference (11.1)

where θ is approximately equal to the angle that the rays make relative to a

perpendicular line joining the slits to the screen.

The condition for constructive interference at the screen is:

d sin = m , m = 0, +1, +2,... (11.2)

3

The condition for destructive interference at the screen is:

d sin = (m +1/2) , m = 0, +1, +2,... (11.3)

Figure 11.2 Constructive and destructive interference.

m = 0 The zeroth order fringe or central maximum.

m = 1 The first order fringe – is the first bright/dark fringe on either

side of the central maximum (there are two first order fringes), and

so on.

Figure 11.3 A geometric construction to describe Young double-slit experiment.

4

- In the case that y , the distance from the interference fringe to the point of

the screen opposite the center of the slits is much less than L ( y < < L ), one can

use the approximate formula.

For small angle ,

sin tan L

y (11.4)

where :

y is the distance from the central maximum on the screen

L is the distance from the slits to the screen.

The distance of the mth bright fringe (ym) from the central maximum on either

side is

d

mLy brightm

)(

For bright fringes (11.5)

The wavelength of the light

mL

dym for m = 1,2,3………

The separation between adjacent bright fringes

d

Lyyy mm

1 (11.6)

For dark fringes :

d

Lmy darkm

)2/1()(

For dark fringes (11.7)

If d < < L then the spacing between the interference can be large even when the

wavelength of the light is very small (as in the case of visible light). This give a

method for (indirectly) measuring the wavelength of light.

The above formulas assume that the slit width is very small compared to the

wavelength of light, so that the slits behave essentially like point sources of

light.

5

Example 1:

Light of wavelength 632.8 nm falls on a double slit and the third order bright

fringe is seen at an angle of 6.50. Find the separation between the double slits.

Solution:

Given =632.8nm=632.8 x 10-9 m, n=3, 3 =6.50 Find d?

From equation : nd sin ,

we have

sin

nd

3sin

3

d

5.6sin

)108.632(3 9

x

mmx 17107.1 5

Example 2:

In a young’s double slit experiment, if the separation between two slits is 0.10mm

and the distance from the slits to a screen is 2.5 m, find the spacing between the

first order and the second order bright fringes for light with the wavelength of

550 nm.

Solution:

Given =550nm=550 x 10-9 m, L=2.5, d= 0.1mm=0.10x10-3m, Find y2- y1

From d

nLy

y2- y1 = d

L)2(-

d

L)1(=

d

L= cm

x

x4.1

1010.0

)10550)(5.2(3

9

Example

Light of wavelength 460nm falls on two slits spaced 0.300 mm apart. What is the

required distance from the slits to a screen if the spacing between the first and

second dark fringes is to be 4.00mm?

Solution:

L = (Δy) d / λ = (4.00 x 10-3 m)(3.00 x 10-4 m) / (460 x 10-9 m) = 2.61 m

6

11.3 Change of phase due to reflection

- Another simple arrangement for producing an interference pattern with a single

light source is known as Llyod’s mirror as shown in figure 8.4.

Figure 11.4: Llyod’s mirror

An interference pattern is produced on a screen at P as a result of the

combination of the direct ray (SP) and the reflected ray. The reflected ray

undergoes a phase change of 1800.

A ray reflecting from a medium of higher refractive index undergoes a 1800

phase change.

A ray reflecting from a medium of lower refractive index undergoes no phase

change.

7

Fig. 11.5

11.4 Interference in thin films

Interference observed in light reflected from a thin film is due to a

combination of rays reflected from the upper and lower surfaces. Refer figure

below.

Interference effects are commonly observed in thin films.

Eg: soap bubbles, thin layers of oil in water.

Figure 11.6 Interference observed in light reflected from a thin film is due to a combination of

rays reflected from the upper and lower surfaces.

8

Consider a film of thickness t and index of refraction n as in the Figure 9.5.

To determine whether the reflected rays interfere constructively or

destructively, note these following facts:

1. An electromagnetic wave traveling from a medium of index of refraction n1

toward a medium of index of refraction n2 undergoes a 1800 phase of change

on reflection when n2> n1. There is no phase change in the reflected wave if

n2< n1.

n1 > n2 0 phase shift

n1 < n2 1800 phase shift

2. The wavelength of light λn in a medium with index of refraction n is given by

n

n

where λis the wavelength of light in vacuum. (11.8)

The condition for constructive interference is

2t = (m + ½ ) λn m = 0,1,2,…. (11.9)

Because λn = λ/ n , the equation will become

2nt = (m + ½ ) λ m = 0,1,2,… (11.10)

The condition for destructive interference is

2nt = mλ m= 0,1,2….. (11.11)

Two factors influence interference:

1. phase reversals on reflection

2. differences in travel distance

11.5 Diffraction

Diffraction is the spreading of light or any other wave as it passes through

openings or around obstacles.

When the wavelength is small compared with the size of obstacles/openings

then diffraction effects are small, and diffraction effects increase as the size

of the obstacles/openings decreases

Similarly, obstacles only generate strong echoes if they are larger than the

wavelength of the waves.

9

Figure 11.7 The light from two slits overlaps as it spreads out and producing interference fringes / Diffraction of waves through one slit

This bending is due to Huygen's principle, which states that all points along a

wave front act as if they were point sources. Thus, when a wave comes against a

barrier with a small opening, all but one of the effective point sources are

blocked, and the light coming through the opening behaves as a single point

source, so that the light emerges in all directions, instead of just passing

straight through the slit.

The diffraction pattern from a single slit of width w consist of a broader

central maximum and some narrower side maxima (the width of the central

maxima is twice that of the side maxima).

-In diffraction, the dark fringes rather than the bright fringes are analyzed.

Figure The Fraunhofer diffraction pattern o a single slit.

10

11.6 Single slit diffraction

Figure 11.9 Fraunhofer Single Slit

The diffraction pattern at the right is taken with a helium-neon laser and a

narrow single slit.

Figure 8.9 shows the position of the minima in a diffraction pattern of a single

slit of width a.

According to Huygen’s Principle, each portion of the slit acts as a source of

waves. Hence, light from one portion of the slit can interfere with light from

another portion and the resultant intensity on the screen depends on the

direction θ.

The condition for the dark fringes is given by

a

m sin for m = +1,+2,+3,…. (11.12)

where is the angle for particular minimum designated by m =1,2,3…… on

either side of the central bright fringe.

(there is no dark fringe correspond to n=0)

For a small angle approximation, the position if the dark fringes from the center

of the central bright fringe on a screen can be calculated from

w

mLym

(11.13)

where L – the distance from the single slit to the screen.

11

The width of the m side maximum is the distance between the mth dark and the

(m+1)th dark or Ym+1 –ym .

It is evident from this equation that

For a given slit width ( w ), the greater the wavelength ( ), the wider the

diffraction pattern ( my ).

For a given wavelength ( ), the narrower the slit width ( w ), the wider the

diffraction pattern( my )

The width of the central maximum is twice the width of the side maxima.

Example

Light of wavelength 632.8 nm is incident on a slit of width 0.200mm. An observing

screen is placed 2.5 m from the slit. Find the width of the central maximum and the

position of the third order dark fringe.

Solution:

Given =632.8 nm=632.8 x 10-9m,

w =0.200 mm=0.200 x 10-3 m, L=2.50m

Find: 2y1 and y3

The central maximum is the region between the first order dark fringes on either

side of this maximum, so its width is simply 2y1.

From w

mLym

,

3

9

110200.0

)108.632)(5.2)(1(1

x

x

w

Ly

= 7.91x10-3 m.

So the width of the central maximum is

2y1 =2(7.91mm) = 15.8 mm.

3

9

310200.0

)108.632)(5.2)(3(3

x

x

w

Ly

= 23.7 mm.

Example

Light of wavelength 550nm is incident on a single slit 0.75mm wide. At what

distance from the slit should a screen be placed if the second dark fringe in the

diffraction pattern is to be 1.7 mm from the center of the screen?

12

Solution :

From w

mLym

We have mx

xx

m

wyL m 2.1

)10550)(2(

)1075.0)(107.1(9

33

Example

Light of wavelength 587.5 nm illuminates a single 0.75 mm wide slit.

a) At what distance from the slit should a screen be placed if the first

minimum in the diffraction pattern is to be 0.85 mm from the central

maximum?

b) Calculate the width of the central maximum.

Solution :

a) At the first dark band: sin θ = λ/a = 5.875 x 10-7 m / 7.5 x 10-4 m = 7.83 x 10-4

But also, sin θ = y/ L , so L = y / sin θ , Thus L = 8.5 x 10-4/ 7.83 x 10-4 = 1.09m

b) The width of central maximum = 2y = 2(0.85mm) 1.70 mm

Example

A screen is placed 50.0 cm from a single slit which is illuminated with light of

wavelength 680 nm. If the distance between the first and the third minima in the

diffraction pattern is 3.00mm, what is the width of the slit?

Solution:

The angles at which a dark fringe can occur are given by

sin θ = mλ/a ,

The screen positions of these dark fringes are:

ym = L tan θ.

Making approximation sin θ = tan θ, gives ym = m L (λ/ a) as the location of the mth

order dark fringe.

The distance from the first to the third dark fringes is then

Δy = y3 – y1 = 2L (λ/a).

With L = 50.0 cm, λ= 680 nm and Δy = 3.00mm, this gives:

3.00 x 10-3 = 2(5.00 x 10-1 m) (6.80 x 10-7m / a ) or a = 0.227 mm.