Name: Constructive and Destructive Interference of Light Waves
Uploaded: 2019-07-19
Duration: 11 min 15 s
Description: If 2 balls rolling on a surface with a certain speed approach each other, then we expect that they will collide, and their velocities get altered after the c...

Question

Explain interference of light and mention essential conditions for it. Derive the formula for fringe width fringe by giving the principle of Young's experiment.

Accepted Answer

Interference is the phenomenon in which two waves superpose to form the resultant wave of the lower, higher or same amplitude.

We can see from the above equation that as the separation d between the slits is increased, the fringe width is decreased. If d becomes much larger than X, the fringe width will be very small. The maxima and minima, in this case, will be so closely spaced that it will look like a uniform intensity pattern. This is an example of the general result that the wave effects are difficult to observe if the wavelength is small compared to the dimensions of the obstructions or openings to the incident wavefront.

We already know that the path difference

$△ x = nλ$ For bright fringes and

$△ x = (n + \frac{1}{2}) \times λ$ For dark fringes

Now let us study the given figure.

Suppose $S_{1} A$ Is the perpendicular from $S_{1}$ to $S_{2} P$ .

Suppose,

$D = OB =$ Separation between slits and the screen,

D = separation between th slits

And D>>d.

Under the above approximation $S_{1} P$ and $S_{2} P$ Are nearly parallel and hence $S_{1} A$ Is very nearly perpendicular to $S_{1} P$ , $S_{2} P$

And OP. AS $S_{1} S_{2 }$ Is perpendicular to OB and $S_{1} A$ Is perpendicular to OP, we have

$∠ S_{2} S_{1} A = ∠ POB = θ$

This is a small angle as D>>d.

The path difference is

$△ x = {PS}_{2} - {PS}_{1} \approx {PS}_{2} - PA$

$= S_{2} A = dsinθ \approx dtanθ$

$= d \times \frac{y}{D} $

The centres of the bright fringes are obtained at distances y from the point B, where

$△ x = d \times \frac{y}{D} = nλ$

Or, $y = dn \frac{λ}{D} $ i.e.,

at $y = 0, \pm \frac{λ}{Dd} , \pm 2 \frac{λ}{Dd}, \pm 3 \frac{λ}{Dd}$ , $. . . . etc$

Similarly, we can show for dark fringes

$y = \pm \frac{Dλ}{2 d} , y = \pm \frac{3 Dλ}{2 d} , y = \pm \frac{5 Dλ}{2 d} , . . . . . . . etc$

The width fringe is therefore,

$ω = d \frac{λ}{D} $

Explain interference of light and mention essential conditions for it. Derive the formula for fringe width fringe by giving the principle of Young's experiment.

Important Questions on Wave Optics

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