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Question

To produce interference fringe pattern, draw a ray diagram of Young's double slit experiment. Derive an expression of fringe width for bright interference fringes. If fringe width of bright fringes is 2mm, write the fringe width for dark fringes. Whether the center point of interference fringe pattern is bright or dark? Explain it.

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Solution

Let the separation between the slits in Young's double slit experiment be d and λ be the wave length of light used.Also, the light reaching at point P via different paths interfere and formed nth bright fringe at point P.From figure, S1P=√S1B2+PB2=√D2+(yn−d2)2Similarly, S2P=√S2A2+PA2=√D2+(yn+d2)2Path difference between the light reaching at P Δx=S2P−S1P∴ Δx= √D2+(yn+d2)2− √D2+(yn−d2)2OR Δx= D[√1+(yn+d2)2D2− √1+(yn−d2)2D2]Using (1+nx)n≈1+nx for x<<1For d<<D, we get Δx= D[1+(yn+d2)22D2]− [1+(yn−d2)22D2]OR Δx= 12D[(yn+d2)2− (yn−d2)2]OR Δx= 12D[(y2n+d24+ynd)2− (y2n+d24)2+ynd]⟹ Δx=yndDFor bright fringe to be obtained, path difference must be equal to integral value of lambda i.e. Δx=nλ∴ yndD=nλ ⟹yn=nλdDFringe width of bright fringe β=yn−yn−1∴ β=nλDd−(n−1)λDd⟹ β=λDdIn Young's double slit experiment, bright and dark fringes of equal width are formed.Fringe width of bright fringe β=2mm (Given)Thus width of dark fringes β′=β=2mmAs point O is equally spaced for both the slits, thus path difference between the lights reaching the point O is zero and hence bright fringe is formed at point O. Thus centre point of interference fringe pattern is bright.

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