Interference of light | Oscillations, Waves & Optics - Physics PDF Download

Introduction

When the two waves of the same frequency travel in approximately the same direction

and have a constant phase difference the resultant intensity of the light is not distributed

uniformly in space. The non uniform distribution of light intensity due to superposition of

two waves is called interference.

At some points the intensity is a maximum and the interference at these points is called

constructive interference at some other points, the intensity is a minimum (possibly even

zero) and the interference at these points is called destructive interference.

Usually when two lights waves are made to interfere, we get alternate dark and bright

bands of a regular or irregular shape.

Young’s double slit Experiment

Purpose of this experiment is to demonstrate the fundamental properties of light

diffraction and interference patters.

Let S₁ and S₂ represent the two pinholes of Young’s interference experiment. We want

to determine the positions of maxima and of minima on line LL′ !which is parallel to the

y axis and lies in the plane containing points S, S₁ and S₂ . We will show that the

interference pattern (around point O) consists of a series of dark and bright lines.

For an arbitrary point P (on line LL′ ) to correspond to a maximum or minima,

S₂P − S₁P = nλ

Now, (S₂P)² - (S₁P)² =

= 2y_nd

⇒ S₂P − S₁P =

If y_n, d << D , then negligible error will be introduced if S₂P + S₁P is replaced by 2D

S₂P − S₁P =

Thus, the dark and bright fringes are equally spaced, and the distance between two

consecutive dark (or bright) fringes is given by

β = y_n+1 - y_{n = λD/d, which is the expression for the fringe width.}

_{The Intensity Distribution}

Let E₁ and E₂ be the electric fields produced at point P by S₁ and S₂ , respectively. The

electric fields E₁ and E₂ will, in general, have different directions and different

magnitudes. However, if the distances S₁P and S₂P are very large in comparison to the

distance S₁S₂ , the two fields will almost be in the same direction. Thus, we may write

Where, ˆi represents the unit vector along the direction of either of the electric fields. The

resultant field is given by

E = E₁ + E_{2 =}

_{The intensity I is proportional to the square of the electric field and is given by I = KE}²

I =  

+

where K is a proportionality constant.

The factor cos (ωt −φ ) will oscillate between +1 and –1, and its average will be zero,

whereas average value of cos² (ωt −φ ) is 1/ 2 .

Thus the intensity, that a detector will record, will be given by

I = I₁ + I₂ + 2√I₁I₂cosδ ,

where δ =  represents the phase difference between the displacements reaching point P from S₁ and S₂

Further I₁ =  and I₂ = 1/2KE²₀₂ represents the intensity produced by source S₁ and from S₂ on the screen.

Condition of Maximum Intensity

The maximum values of cosδ is +1. Therefore the maximum values of I is given by

I_max =

The maximum intensity occurs when, Phase difference: δ = 2nπ ; n = 0,1,2,......

or Path difference: S₂P − S₁P = nλ

Condition of Minimum Intensity

The minimum values of cosδ is −1, as a result the minimum values of I is given by

The minimum intensity occurs when Phase difference: δ = (2n +1)π ; n = 0,1, 2,......

or Path difference: S₂P − S₁P =

Intensity when S₁P and S₂P are extremely large

If the distances S₁P and S₂P are extremely large in comparison to d , then I₁ ≅ I₂ = I₀

and

I = 2I₀ + 2I₀ cosδ = 4I₀ cos² δ / 2

The intensity distribution is shown in below

Conditions for sustained Interference

• The two interfering waves should be coherent
• The two waves should be of same frequency
• If the interfering waves are polarized, they must be in the same state of polarization.

Conditions for observation

• The spacing between the light sources (d) should be as small as possible.
• The distance (D)of the screen from the two sources should be quite large.

Conditions for good contrast

• The amplitudes of the interfering waves should be equal or at least very nearly equal.
• The two sources must be narrow.
• The two sources should give monochromatic or very nearly monochromatic light, or else the path difference should be very small.

Example 1: Suppose in the double-slit arrangement, d = 0.150mm , D = 120cm, λ = 833nm, and y = 2.00cm.
(a) What is the path difference δ for the rays from the two slits arriving at point P? 
(b) Express this path difference in terms of λ
(c) Does point P correspond to maximum, a minimum, or an intermediate condition?

(a) The path difference is given byδ = d sinθ . When D >> y,θ is small and we can make the approximation sinθ ≈ tanθ =  y/D. Thus,

= 2.50 x 10^-6m

(b) From the answer in part (a), we have  or δ = 3.00λ

(c) Since the path difference is an integer multiple of the wavelength the intensity at point

P is a maximum.

Example 2: In Young’s double-slit experiment, suppose the separation between the two slits is d = 0.320 mm. If a beam of 500 nm light strike the slits and produces an interference pattern. How many maxima will there be in the angular range −45.0^o <θ < 45.0^o ?

On the viewing screen, light intensity is a maximum when the two waves

interfere constructively. This occurs when

d sinθ = nλ,       n = 0, ±1, ±2...

where λ is the wavelength of the light. Atθ = 45.0^o , d = 3.20×10⁻⁴m and λ = 500×10⁻⁹m, we get n = = 452.5

Thus, there are 452 maxima in the range0 <θ < 45.5^o . By symmetry, there are also 452 maxima in the range −45.0^o < θ < 0 . Including the one for n = 0 straight ahead, the total number of maxima is N = 452 + 452 + 1 = 905

Example 3: Let the intensity on the screen at a point P in a double slit interference pattern be 60.0% of the maximum value
(a) What is the minimum phase difference (in radians) between sources?
(b) In (a), what is the corresponding path difference if the wavelength of light is λ = 500nm?

(a) The intensity I=  where I_max is the maximum light intensity.

Thus, 0.60 cos² ⇒δ = 2cos⁻¹(√0.60) = 78.5^o =1.37 rad

(b) The phase difference δ is related to the path difference Δ and the wavelength λ by

Δ == = 109nm

Intensity of Three Slit Interference

Suppose a monochromatic coherent source of light passes through three parallel slits,

each separated by a distance d from its neighbor, as shown in following Figure

The waves have the same amplitude E₀ and angular frequencyω , but a constant phase

difference δ =

Example 4: Show that the intensity is I = where I₀ is the maximum intensity associated with the primary maxima.
(b) What is the ratio of the intensities of the primary and secondary maxima?

(a) Let the three waves emerging from the slits be

E₁ = E₀ sinωt, E₂ = E₀ sin ω t +δ, E₃ = E₀ sin ω t + 2δ

Using the trigonometric identity sinα + sinβ =

The sum of E₁ and E₃ is

E₁ + E₃ = E₀[sinω t + sin ω t + 2δ] = 2E₀ cosδ sin (ωt + δ)

The total electric field at the point P on the screen is

E = E₁ + E₂ + E₃ = 2E₀ cosδ sin(ωt +δ) + E₀ sin (ωt +δ) = E₀(1+ 2cosδ) sin (ωt + δ)

where δ =

The intensity is proportional to E² , thus

=

∵

The maximum intensity I₀ is attained when cosδ =1, thus I_0 =

_Thus,

_{(b) The interference pattern is shown in following figure}

_{Form the figure, we see that the minimum intensity is zero, and occurs when cosδ = -1/2}

_{The condition for primary maxima is cosδ = +1, which gives}

In addition, there are also secondary maxima which are located at cosδ = −1. The

condition implies

δ = (2m+ 1)π

⇒

The intensity ratio is

Example 5: Consider the interference pattern from the three slits illustrated (shown in figure). Assume the openings of the individual slits are the same .At what value of θ is the first principal maximum? (i.e., the wavelets from all three slits are in phase).

The electric field E(θ ) on the screen is the sum of the fields produced by the slits individually:

where

Then the total intensity at θ is

For θ = 0 , we have I (0) ~ 9A² .

The expression for I (θ ) shows that the first principal maximum occurs at δ = 2π , i.e.,

Displacement of Fringes by the introduction of a thin Lamina Sheet

When a thin transparent plate, say of glass or mica, is introduced in the path of one of the

two interfering beams, the entire fringe-pattern is displaced to a point towards the beam

in the path of which the plate is introduced. Let a thin plate of thickness t be introduced in the path of light from S₁ . Let μ be the refractive index of the plate for the monochromatic light employed. Time taken for the journey from S₁ to P

v₀ = velocity of light in air &

v₁ = velocity of light in plate

Hence the effective path in air from S₁ to P is

At any point P , effective path difference

=

For bright fringe

In the absence of the plate t = 0 ;

Thus the displacement of n^th fringe

This expression is independent of n so that the displacement of any dark fringe is also

What Are Interference Fringes?

• As laser light is diffracted through the two barrier slits, each diffracted wave meets the other in a series of steps, as illustrated in Figure 4 (and graphically in the interactive tutorial described above). 
• Sometimes the waves meet in step (or in phase; constructive interference), sometimes they meet out of step (or out of phase; destructive interference), and sometimes they meet partially in step. 
• When the waves meet in step, they add together owing to constructive interference, and a bright area is displayed on the screen. 
• In areas where the waves meet totally out of step, they will subtract from each other owning to destructive interference, and a dark area will appear in that portion of the screen. 
• The resulting patterns on the screen, a product of interference between the two diffracted beams of laser light, are often referred to as interference fringes.

• Other types of experiments have been devised to demonstrate the wave-like nature of light and interference effects. 
• Most notable are the single mirror experiment of Humphrey Lloyd and the double mirror and biprism experiments devised by Augustin Fresnel.

Newton’s Rings Experiment

• Sir Isaac Newton, the famous 17th century mathematician and physicist, was one of the first scientists to study interference phenomena. 
• In his famous Newton's Rings experiment, he placed a convex lens of large curvature radius on a flat glass plate and applied pressure to hold the lens and glass plate together. 
• When he viewed the plates through reflected sunlight, he observed a series of concentric light and dark highly colored bands of light similar to those illustrated in Figure 5. 
• Newton recognized that the rings indicated the presence of some degree of periodicity and used this observation to suggest a wave theory of light. 
• Despite this, Newton regarded light as a stream of particles.
• The rings occur because of a thin layer of air that exists between the curved convex and flat glass surfaces. 
• Light reflected from the top and bottom surfaces of the glass is superimposed (combined) and produces interference patterns that appear as colored rings. 
• This principle is often used by lens manufacturers to test the uniformity of large polished surfaces.