Question

Explain Young's double slit experiment set up and obtain the equation for the path difference.

Accepted Answer

Young’s double slit experiment demonstrated the phenomenon of interference of light. Consider two fine slits $S 1 and S_{2}$ at a small distance d apart. Let the slits be illuminated by a monochromatic source of light of wavelength $λ$ These source will be coherent, since they originate from the same source . Let GG′ be a screen kept at a distance D from the slits. The two waves emanating from slits $S 1 and S_{2}$ superimpose on each other resulting in the formation of an interference pattern on the screen placed parallel to the slits.

Let O be the centre of the distance between the slits. The intensity of light at a point on the screen will depend on the path difference between the two waves reaching that point. Consider an arbitrary point P at a distance x from O on the screen.

Path difference between two waves at P = S₂P − S₁P

On the screen, the intensity at point P is maximum or minimum as the path difference from the two slits is an integral multiple of wave length of an odd multiple of half-wave length

For maxima ${(S_{1} P)}^{2} - {(S_{2} P)}^{2} = D^{2} + {[x + \frac{d}{2}]}^{2} - D^{2} + {[x - \frac{d}{2}]}^{2} = 2 x d$

$S_{2} P - S_{1} P = \frac{2 xd}{S_{2} P + S_{1} P}$ as $d < < < D$ $S_{2} P + S_{1} P = 2 D (S_{2} P = S_{1} P = d)$

Hence, the path difference = $S_{2} P - S_{1} P = \frac{xd}{D}$

Explain Young's double slit experiment set up and obtain the equation for the path difference.

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