Question

Draw a neat labelled diagram of Young’s Double Slit experiment. Show that β =λD/d , where the terms have their usual meaning (either for bright or dark fringe).

Answer 1

Consider a point P on the viewing screen; the screen is located at a perpendicular distance D from the screen containing the slits S₁ and S₂, which are separated by a distance das shown in figure. Let us assume that the source of light is monochromatic having a wavelength λ. Under these conditions, the waves emerging from S₁ and S₂ have the same frequency and amplitude and are in phase.

The light intensity on the screen at P is the resultant of the lights coming from both the slits. As seen in the diagram a wave from the lower slit travels farther than a wave from the upper slit. This difference in path is called the path difference. Therefore, if ‘x’ is the path difference between the waves from slits S₂ and S₁ at point P, then

x = r₂ - r₁ = dsinθ  .....(i)

where it is assumed that r₁ and r₂ are parallel, which is approximately true because D is much greater than d. The value of this path difference determines whether or not the two waves are in phase, when they arrive at point P.

If the path difference is either zero or some integral multiple of the wavelength, the waves are in phase at P and constructive interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by

x = d sinθ = nλ  ..............(2)

In order to find the position of maxima measured vertically from O to P i.e., y we assume that D >> d. Under these conditions, θ is small, and so the approximation sin θ ~ tan θ. Therefore, from triangle POQ in figure, we find that

sin θ = tan θ = \(\frac{Y}{D}\)  .....(3)

Substituting this in equation (2), we have

\(d\frac {y}{D}\) = nλ    ....(4)

Rewriting the above equation, we have

y = \(\frac{nD\lambda}{d}\)  .....(5)

This gives the distance of the nth maxima

 from the center O of the Screen. For n = 0,1,2,3,4......... the values of y are

y₀ = 0  .... position of the central bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. therefore

y_n - y_n-1 = \(\frac{nD\lambda}{d}\) - \(\frac{(n-1)D\lambda}{d}\) = \(\frac{D\lambda}{d}\) .....(6)

This gives the width of the dark fringe.

Similarly, when he path difference is av odd multiple of \(\frac{\lambda}{2}\), the waves arriving at P will be 180° out of phase and will give rise to destructive interference. Therefore, the condition for dark fringes, or destructive interference, at P is given by

x = d sinθ = (2n - 1)\(\frac{\lambda}{2}\) ...(7)

Using equation (2), we have

This gives the distance of the nth minima from he centre O of the screen. For n is equal to 1,2,3, ...... we have

Now, the spacing between two consecutive dark fringes the  width of the bright fringe, therefore, width of the bright fringe is

From equations (6) and (10), we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by