What is interference of light? Find an expression for fringe widt… - Vidyadip

Question

What is interference of light? Find an expression for fringe width in 'Young's double slit experiment.

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Answer

When two waves of light having same frequency and same ampliitude propagate in a medium simultaneously in the same direction then due to their superposition the resultant intensity in the region of superposition is, in general, different from the sum of intensities due to each wave separately. This modification in the distribution of intensity of light in region of superposition is called interference. In the region of superposition there are certain points at which waves superimpose in such a way that resultant intensity is greater than the sum of intensities due to separate waves. Interference at such points is called constructive interference. At other points in the region of superposition the resultant intensity is less than the sum of intensities due to separate waves and interference at these points is called as destructive interference.
Expression for Fringe Width : In Fig. S is a narrow slit illuminated by monochromatic light. S₁ and S₂ are two parallel narrow slits placed very close to each other. Each of S₁ and S₂ is at equal distance from S. The light waves from S reach at S₁ and S₂ in the same phase.

According to Huygen's principle, S₁ and S₂ act as new light-sources from which secondary wavelets originate. These wavelets form interference fringes on a screen XY placed at a distance D from double slit arrangement. S₁S₂, i.e., CO = D. Suppose the distance between the two slits is d i.e., S₁S₂ = d.
Let the right bisector CO of distance S₁S₂ meet the screen XY at a point. D and let P be a point on the screen at a distance x from O, where both waves starting from S₁ and S₂ reach simultaneously, traversing different distances S₁P and S₂P. Hence the path difference between them is $\left( S _2 P - S _1 P \right)$. Let a perpendicular S₁A be drawn from S₁ on S₂P. Then S₂A will be equal to the path difference between two waves. Therefore,
$\left( S _2 P - S _1 P \right)= S _2 A$.
The right angled triangles $\triangle S _1 AS _2$ and $\triangle POC$ are similar. Therefore
$\frac{ S _2 A}{ S _1 S_2}=\frac{ OP }{ CP }$
The distance CO is large in comparison to S₁S₂ and so CP may be taken equal to CO.
$\therefore \quad \frac{ S _2 A}{ S _1 S_2}=\frac{ OP }{ CO } \Rightarrow \frac{ S _2 A}{d}=\frac{ x }{ D }$
$\therefore \quad S _2 A=\frac{ x d }{ D }$
i.e. Path difference $=\frac{x d}{D}$
Now, according to the value of path difference bright or dark fringes will be obtained at P.
Locations of Bright Fringes : Bright fringes will be obtained at those points on the screen where the path difference between two interfering light waves will be 0, $\lambda, 2 \lambda, \ldots$ ... Hence for constructive interference i.e. bright fringe, we have
Path difference $\frac{x d}{ D }=m \lambda$
(where $m=0,1,2,3, \ldots$ )
$\Rightarrow \quad x = m \left(\frac{ D \lambda}{ d }\right)$ ...(1)
Thus equation (1) gives the location of mth bright fringe.
Hence the location of various bright fringes will be as follows :
For $m=0$; $x_0=0$
which is the position of central bright fringe which is called zero order bright fringe also.
For m = 1 ; $x_1= D \lambda / d$
which gives the location of first bright fringe.
For m = 2 ; $x_2=2 D \lambda / d$
which gives the location of second bright fringe and so on.
Location of Dark Fringes : Dark fringes will be obtained at those points on the screen where the path difference between two interfering light waves will be $\lambda /$ $2,3 \lambda / 2,5 \lambda / 2, \ldots$ ... Hence for destructive interference i.e., dark fringe, we have
Path difference : $\frac{x d}{ D }=(2 m-1) \frac{\lambda}{2}$
(where $m=1,2,3, \ldots$.)
$\Rightarrow \quad x=\frac{(2 m-1) D \lambda}{2 d}$ ...(2)
Thus, equation (2) gives the location of mth dark fringe.
Hence the locations of various dark fringes will be as follows :
For $m=1$ ; $x_1=\frac{ D \lambda}{2 d}$,
which gives the location of first dark fringe.
For $m=2$; $x_2=\frac{3 D \lambda}{2 d}$.
which gives the location of second dark fringe.
For $m=3$; $x_3=\frac{5 D \lambda}{2 d} ;$
which gives the location of third dark fringe and so on.
From the above discussion about the location of bright and dark fringes it is obvious that the bright and dark fringes are located alternately in interference pattern.
Fringe Width : Since it is equal to the distance between two consecutive bright or dark fringes, hence if $x_n$ and $x_{n+1}$ be the distances of the $n$th and $(n+1)$ th fringes respectively then fringe width :
$W =x_{n+1}-x_n$
For mth bright fringe :
$x_m=\frac{m D \lambda}{d}$
$\Rightarrow \quad x_n=\frac{n D \lambda}{d}$ and $x_{n+1}=\frac{(n+1) D \lambda}{d}$
$\therefore \quad W =x_{n+1}-x_n$
$\Rightarrow \quad W =\frac{(n+1) D \lambda}{d}-\frac{n D \lambda}{d}$
$\Rightarrow \quad W =\frac{ D \lambda}{ d }$ ...(3)
For mth dark fringe :
$x_m=\frac{(2 m-1) D \lambda}{d}$
$\Rightarrow \quad x_n=\frac{(2 n-1) D \lambda}{d}$
and $x_{n+1}=\frac{\{2(n+1)-1\} D \lambda}{2 d}$
$\Rightarrow \quad x_{n+1}=\frac{(2 n+1) D \lambda}{2 d}$
$\therefore \quad W =x_{n+1}-x_n$
$\Rightarrow \quad W =\left[\frac{(2 n+1) D \lambda}{2 d}\right]-\left[\frac{(2 n-1) D \lambda}{2 d}\right]$
$\Rightarrow \quad W =\frac{ D \lambda}{ d }$ ...(4)
From equations (3) and (4) it is obvious that the bright and dark fringes are of equal fringe width.
In this way in interference pattern obtained on the screen bright and dark fringes are equally spaced in alternate order.

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