Coherent & in-coherent source of light Young's Double slit Experiment

COHERENT AND INCOHERENT SOURCE OF LIGHT:

Two source of light is said to be coherent if they have same frequency or wavelength, same amplitude or nearly equal amplitude and same phase difference or nearly constant phase difference.

All other sources of light which does not meet the restriction are incoherent. Two independent sources of light can not be coherent , because their phase relationship alters in random orders. So that it becomes unpredictable even though they may have same frequency or wavelength, same amplitude or nearly same amplitude and same phase at a point. Coherent source can be produced in two ways.

Interference by division of wavefront:

Wavelets from different parts of same wave front are made to travel along different path only to recombine to give rise to an interference patterns . since wavelets originate from the same wave fronts, condition of coherent source is satisfied. A narrrow source is required in such cases. Young's double slit experiment, Fresnel's double slit bi-prism and Lloyd's single mirror belongs to this category. In this class of interference, we do not obtain complementary patterns.

• Division of wavefront

\(\rightarrow\) Young's double slit experiment

\(\rightarrow\) Lloyd's single mirror experiment

\(\rightarrow\) Fresnel's Bi-prism

• Divission of amplitude: In this class of interference, the amplitude of a wave front gets divided into two parts. The two parts follow different paths and recombine together to produce interference. A broader or extended source is required in such cases. Interference in thin films and Newton's ring belongs to this category. in this class of interference, we obtain complementary patterns.

\(\rightarrow\) Newton's ring experiment

\(\rightarrow\) Michelson's Interferometer

\(\rightarrow\) Fabry-Perot Interferometer

YOUNG'S DOUBLE SLIT EXPERIMENT:

Young's performed an experiment for the demonstration of wave nature of light. The points \(S_1\) and \(S_2\) are two coherent sources produced by the division of the wave front. Let any point P on the screen which is x distance from the foot of perpendicular drawn from the source on the screen. The separation between the slit \(S_1\) and \(S_2\) is called slit separation and it is denoted by d. The distance between slit and the screen is D.

Let us draw two perpendicular from the point sources \(S_1\) and \(S_2\) on the screen as the point E and F.

Here, we have to find the path difference between the light waves coming from \(S_1\) and \(S_2\) at the point P.

From triangle \(S_2\)FP,

$$(S_1P)^2=(S_1F)^2+(FP)^2$$

$$=D^2+(x+\frac{d}{2})^2\dotsm(1)$$

From the triangle \(S_1\)EP,

$$(S_1P)^2=(S_1M)^2+(EP)^2$$

$$=D^2+(x-\frac{d}{2})^2\dotsm(2)$$

Now, subtracting the equation (2) from (1), we get

$$(S_2P)^2-(S_1P)^2=D^2+( x+\frac{d}{2})^2-D^2-(x-\frac{d}{2})^2$$

$$or,\;\;\;S_2P-S_1P=\frac{2xd}{S_2P+S_1P}\dotsm(3)$$

In the real experiment, the point P lies just above the point C and the separation between the slit is very small. Therefore \(S_2\)P tends to D and \(S_1\)P tends to D.

\(\therefore\) The equation (3) is modified as $$S_2P-S_1P=\frac{2xd}{D+D}=\frac{xd}{D}$$

\(\therefore\) \(S_2P-S_1\)P = \(\frac{xd}{D}\dotsm(4)\)

SPECIAL CASE;

• Condition of bright fringes;

The bright fringes are produced when the path difference appears in the orders of n\(\lambda\), i.e.

$$\frac{xd}{D}=n \lambda$$

$$x=\frac{n\lambda D}{d}$$

So,

$$(x_1)_{bright}= \frac{\lambda D}{d}$$

$$(x_2)_{bright}= \frac{2 \lambda D}{d}$$

$$(x_n)_{bright}= \frac{n \lambda D}{d}$$

$$(x_{n+1})_{bright}= \frac{(n+1) \lambda D}{d}$$

$$\therefore\;\;\;(x_{n+1})_{bright}-(x_n)_{bright}=\frac{(n+1) \lambda D}{d}-\frac{n \lambda D}{d}$$

$$=\frac{\lambda D}{d}\dotsm(5)$$

(ii) Dark fringes are produced when the path difference appears in the order of (2n-1)\(\frac{\lambda}{2}\), i.e.

$$\frac{x d}{D}=(2n-1)\frac{\lambda}{2}$$

$$or,\;\;\;x=\frac{(2n-1)}{2d}\lambda D$$

SO, \((x_1)_{dark}=\frac{\lambda D}{2d}\)

$$(x_2)_{dark}=\frac{3 \lambda D}{2d}$$

$$(x_n)_{dark}=(2n-1)\frac{\lambda D}{2d}$$

$$(x_{n+1})_{dark}=(2n+1)\frac{\lambda D}{2d}$$

So, \((x_{n+1})_{dark}-(x_n)_{dark}=(2n+1)\frac{\lambda D}{2d}-(2n-1)\frac{\lambda D}{2d}\)

$$(x_{n+1})_{dark}-(x_n)_{dark} =\frac{\lambda D}{2d}\dotsm(6)$$

Fringe Width(\(\beta\)) :

The separation between two successive bright fringes or dark fringes is called fringe width. It is denoted by \(\beta\). In the Young's double slit experiment, the fringe width is given by;

$$\beta=\frac{\lambda D}{d}$$

Determination of refractive index ( or thickness) of any thin transparent medium:

The schematic diagram for this is same as given in the Young's double slit experiment shown in the above.

A thin medium having refractive index and the thickness (t) is introduced along the path of one of the interfering beams in Young's double slit experiment. The time taken by the light to reach the point p is adjusted same for both waves. So, the time taken by light to reach from the point \(S_2\) to the point P is given by

$$Time=\frac{S_2 P}{c}\dotsm(1)$

Similarly, the time taken by light to reach from the point \(S_1\) to the point P is

$$Time=\frac{S_1 P-t}{c}+\frac{t}{v}\dotsm(2)$$

Here, \(\frac{S_2 P}{c}=\frac{S_1 P-t}{c}+\frac{t}{v}\)

Or, \(S_2 P = ( S_1 P – t ) + \mu t\)

Or, \(S_2 P – S_1 P = \mu t – t\dotsm(3)\)

Let n be the number of fringes displaced due to the new medium, then,

$$(\mu t )t=n\lambda\dotsm(4)$$

Also, we have, x = \(\frac{n\lambda D}{d}\)

Or, \(n\lambda = \frac{y d}{D}\)

$$\therefore (\mu t)t=\frac{y d}{D}\dotsm(5)$$

Equation (4) and (5) are to determine the refractive index of thin medium if the thickness of the medium is provided.

References:

Adhikari, P.B, Daya Nidhi Chhatkuli and Iswar Prasad Koirala. A Textbook of Physics. Vol. II. Kathmandu: Sukunda Pustak Bhawan, 2012.

Jenkins, F.A and H.E White. Fundamental of optics. New York (USA): McGraw-Hill Book Co, 1976.

wood, R.W. Physical Optics. New York (USA): Dover Publication , 1934.