Path Difference, Phase Difference and Young's Double Slit Experiment

Path Difference and Phase Difference

When a wave passes through a medium, the particles of the medium vibrate. When the particles completer one to and fro motion, the wave advances by a distance equal to its wavelength λ. For a complete wave, the wavelength varies in λ and the phase is changed through 2p.

Let there be two waves with a path difference of λ. Then, the phase difference them will be 2p. If the path difference is x, then path difference \( = \frac {2\pi }{\lambda } \times x. \)

$$ \text {hence, phase difference} = \frac {2\pi }{\lambda } \times \text {path difference} $$

Optical Path

The product of the distance travelled by the light in a medium and the refractive index of that medium is called the optical path. If d be the distance travelled by the light in a medium of refractive index µ, then by definition,

$$ \text {Optical Path} = \mu d $$

Let us consider an optically denser medium in which the light travels with a velocity v. if the distance travelled by the light in the medium is d, then the time taken to ravel this distance is given by

$$ t = \frac dv \dots (i) $$

In this time, the light travels a distance L in free space. Then,

$$ L = ct \dots (ii) $$

Now, the refractive index of the medium is given by

$$ \mu = \frac cv \dots (iii) $$

where c is the velocity of the light in a free space. Substituting the value of t from equation (i) in equation (ii), we get

\begin{align*} L = c \times \frac dv = \frac {cv}{v} \\ \text {or,} \: L = \mu d \\ \end{align*}

This is called the optical path.

Conditions for Sustained Interference of Light

1. The two sources of light must be coherent.
2. The amplitude of waves from the two sources should be equal.
3. Two sources should be monochromatic. Otherwise, the fringes of different colours will overlap.
4. The coherent sources must be very close to each other.
5. The two sources should be point sources or very narrow sources.
6. The wave from the sources be in same phase or maintain a constant phase difference with time.
7. The interfering beams should be of same wavelength and frequency.

Young’s Double Slit Experiment

S is a narrow vertical slit (of width about 1 mm) illuminated by a monochromatic source of light. At a suitable distance (about 10 cm ) from S, there are two fine slits S₁ and S₂ about 0.5 mm apart at equidistant from S. when a screen is placed at a larger (about 2m) from the slits S₁ and S₂, alternate bright and dark bands appear on the screen. The appearance of bright and dark bands are called the fringes.

Theory of Interference of Light

Suppose S₁ and S₂ be two fine slits at a small distance d apart in the figure. Let slits are illuminated by monochromatic light from a strong source S of wavelength λ and MN is a screen at a distance D from the double slits. The two waves starting from S₁ and S₂ superimpose upon each other resulting an interference pattern on the screen placed parallel to the double slit as in the figure.

Let O be the centre between the slits S₁ and S₂. Draw S₁P, S₂P and OC perpendicular to MN. The intensity of light at a point on the screen will depend upon the path difference between the two waves arriving at the point. The point C on the screen lies on the perpendicular bisector of S₁ and S₂. Therefore, the path difference between two waves reaching C is zero and hence, they are in phase. So, the point C is the position of maximum intensity. It is called central maximum.

Consider a point P at a distance x from C. The path difference between two waves arriving at P is given by

$$\text {path difference} = S_2P – S_1P $$

From the geometry in figure, it is found that

\begin{align*} PQ = x-\frac d2 ; PR = x + \frac d2 \\ \text {And} (S_2P)^2 –(S_1P)^2 = \left [ D^2 + \left ( x + \frac d2 \right )^2 \right ] - \left [ D^2 + \left ( x - \frac d2 \right )^2 \right ]\\ \text {or,} \: (S_2P – S_P)(S_2P + S_1P) = 2xd \\ \text {or,} \: (S_2P – S_P)= \frac {2xd}{BP + AP} \\ \text {In practice, point P lies very close to C.} \\ \text {So} \: S_2P \approx S_1P \approx D \\ S_2P + S_1P = D + D = 2D \\ \text {Path difference} = S_2P – S_1P = \frac {2xd}{2D} = \frac {xd}{D} \\ \end{align*}

The waves from S₁ and S₂ arriving at a point on the screen will interfere constructively or destructively depending upon this path difference. The phase difference for this path difference is given by

$$ \text {Phase difference,} \: \phi = \frac {2\pi}{\lambda } \left (\frac {xd}{D} \right )$$

1. Bright fringes
   If the path difference is an integral is an integral multiple of wavelength λ, then point P is bright. Therefore, for bright fringes
   \begin{align*} \frac {xd}{D} &= n\lambda \\ x &= n\lambda \frac Dd \dots (i) \\ \end{align*}
   where n = 0, 1, 2, 3, … The distance of the various bright fringes from the central maximum at C can be found as follows:
   \begin{align*} \text {For} \: n=0, x_0 = 0 \dots \text {central bright fringes} \\ \text {For} \: n=1, x_1 = \frac {\lambda D}{d} \dots \text {first bright fringes}\\ \text {For} \: n=2, x_2 = \frac {2\lambda D}{d} \dots \text {second bright fringes} \\ \text {For} \: n=n, x_n = n \lambda \frac {D}{d} \dots n^{th} \text { bright fringes} \\ \end{align*}
   The distance between any two consecutive bright fringes is called fringe width, denoted by β.
   \begin{align*} \text {Fringe width,} \: \beta = x_2 – x_1 = \frac {2\lambda D}{d} - \frac {\lambda D}{d} = \frac {\lambda D}{d} \\ \therefore \beta = \frac {\lambda D}{d} \dots (ii) \end{align*}
2. Dark fringes
   If the path difference is an odd integral multiple of half wavelength λ, then point P is dark. Therefore, for dark fringes;
   \begin{align*} \frac {xd}{D} &= (2n-1)\frac {\lambda }{2} \: \text {where} \: n = 1,2,3, \dots \\ \text {or,}\: x &= (2n-1)\frac {\lambda D}{2d} \dots (iii) \\ \end{align*}
   Equation (iii) gives the distance of the dark fringes from point C. the distance of the various dark fringes from point C can be calculated as below:
   \begin{align*} \text {For} \: n=1, x_1 = \frac {\lambda D}{2d} \dots \text {first dark fringe}\\ \text {For} \: n=2, x_2 = \frac {3\lambda D}{2d} \dots \text {second dark fringes} \\ \text {For} \: n=n, x_n = (2n-1)\lambda \frac {\lambda D}{2d} \dots n^{th} \text { dark fringes} \\ \end{align*}
   The distance between any two consecutive dark fringes is called fringe width β, given as
   \begin{align*} \text {Fringe width,} \: \beta = x_2 – x_1 = \frac {3\lambda D}{2d} - \frac {\lambda D}{2d} = \frac {\lambda D}{d} \\ \therefore \beta = \frac {\lambda D}{d} \dots (iv)\\ \end{align*}
   from equation (ii) and (iv), it is clear that width of the bright is equal to the width of the dark fringe. From these two equations it is clear that fringe width increases as the
   1. Wavelength increases.
   2. Distance D of the screen from the sources increases
   3. Distance between the sources decreases.

Reference

Manu Kumar Khatry, Manoj Kumar Thapa,et al. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.