The set up used by the Young to demonstrate the phenomenon of interference of light is as shown in the following Fig.:
S is a narrow slit. The width of the slit is about 1 mm. At a suitable distance from S there are two narrow slits say A and B about 0.5 mm apart. These two slits are placed parallel to S.
There is a screen placed at a large distance of about 2 metre from slits A and B. When we illuminate the slit Shy a monochromatic source of light, alternate bright and dark bands appear on the screen. These bands are called interference bands or interference fringes. The bands disappear, when one of the slits A or B is closed. Now, according to Huygen.s principle when slits .S. is. illuminated by monochromatic source of light, it emits spherical wavefronts and also slits A and B emit spherical wavelets. The full line semicircles represent crests and dotted line semicircles represent troughs. At points, where crests of one wave fall on the crests of other, the interfence is constructive and resultant intensity is maximum. At points, where crests of one wave fall on troughs of the other wave, the interference is destructive and resultant intensity is minimum. Hence, at these points, we observe dark bands. The points D and Fcorrespond to dark bands.
Dark and Bright bands in the interference pattern are caiJed interference fringes. Expression for fringe width . Consider two coherent sources `S_(1)` and `S_2` separated by a distance d. Let Z be the distance between th« screen and the plane of slits `S_(1)` and `S_(3)` (Fig.).
Light waves emitted from 5J and reach point O on the screen after travelling equal distances. So path difference and hence phase difference between these waves is zero. Therefore, they meet at O in phase and hence constructive interference takes place at O. Thus O is the position of the central bright fringe.
Let the waves emitted by `S_(1)` and `S_(2)` meet at point Pan the screen at a distance y from the central bright fringe. The path difference between these waves at P given by:
`Deltax = S_(2)P -S_(1)P`....(i)
From right aangled trianagle, `S_(2)BP`,
`S_(2)P =[S_(2)B^(2) + PB^(2)]^(1//2) =[D^(2) + (y +d/2)^(2)]^(1//2) = D[1+ ((y+{d/2)^(2))/D^(2))]^(1//2)`
Using Binomial theorem, and neglecting terms containing higher powers,
`therefore S_(2)P = D +((y+D/2)^(2))/(2D)`..........(ii)
`S_(1)P = D+((y-d/2)^(2))/(2D)`
Substituting the values of (ii) and (iii) in (i), we get,
`Delta x = D + (y+d/2)^(2)/(2D) -D -(y-d/2)^(2)/(2D)`
`Deltax = (yd)/D`
Bright Fringes {Constructive Interference/Maxima)
If path difference is an integral multiple of `lambda`, then bright fringe will be formed at P
i.e., `(yd)/D = nlambda` or `y =(nlambdaD)/d`
which is the position of irth bright fringe from the central bright fringe.
If `n=1, y_(1) =(lambdaD)/d`, position of first bright fringe from the central bright fringe
If `n=2, y_(2) = (2lambdaD)/d`, position of second bright fringe from the central bright fringe and so , on
If `n=(n-1), y_(n-1) =((n-1)lambdaD)/d`
which is the position of (n- 1)th bright fringe from the central bright fringe. Fringe Width (`beta`). The distance between any two successive bright fringes is called fringe, width.
i.e., `beta = y_(2) - y_(1) =(2lambdaD)/d -(lambdaD)/d`
or `beta = (lambdaD)/d`........(iv)
Dark Fringes (Destructive Interference/Minima) ,
If path difference is odd multiple of `lambda//2` then dark fringe is formed at P.
i.e. `(yd)/D = (2n+1)lambda/2`, where n =1,2,....
`y = ((2n+1)lambdaD)/(2d)`
which is the position of nth dark fringe from.the central bright fringe.
If `n=0, y_(0) = (lambdaD)/(2d)`, Position of first dark fringe from the central bright fringe.
If `n=1, y_(1) = (3lambdaD)/(2d)`, Position of second dark fringe from the central bright fringe.
If `n=(n-1), y_(n-1) =((2n-1)lambdaD)/(2d)`, Position of (n-1)th dark fringe from the central bright fringe. The distance between any two successive dark fringes is aiso called fringe width.
`beta = lambda_(1) - lambda_(0) = (3lambdaD)/(2d) -(lambdaD)/(2d)` or `beta = (lambdaD)/d`.....(iv)
From above, if is clear that fringe widths of bright and dark fringes are equal. Factors on which fringe width depends :
We know, `beta = (lambdaD)/d`
(i) Fringe width (`beta`) is directly proportional to wavelength of light. `bet prop lambda`
(ii) Fringe width (`beta`) is directly proportional to the distance of screen from slits.` beta prop D`
(iii) Fringe width (`beta`) is directly proportional to the distance between the slits.
`beta prop 1/d`
We can increase the fringe width by increasing `lambda` and D and by decreasing d. Special case L If distance between the two slits approaches zero then d=0
and `beta =(lambdaD)/0 = infty`
So no fringe pattern will be visible on the screen..
Special case II. If whole of the apparatus is dipped in water
Wavelength of light in water = `lambda. = lambda/mu`
where `mu`= Refractive index of water
New fringe width (in water) = `beta. =(lambda.D)/d`
`beta. =(lambda)/mu D/d =1/mu (lambdaD)/d = beta/mu`
So in water, fringe width decreases `mu` times.
