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Question

# (a) In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width. (b) The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9:25. Find the ratio of the widths of the two slits.

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Solution

## (a) In Young's double slit experiment, the wavefronts from the two illuminated slits superpose on the screen. This results in formation of alternate bright and dark fringes because of constructive and destructive interference, respectively. The intensity of light it maximum at the centre C of the screen and it is called central maxima. Let S1 and S2 be two slits separated by a distance d. GG' is the screen at a distance D from the slits S1 and S2. Both the slits are equidistant from point C. The intensity of light will be maximum at this point due to the path difference of the waves reaching this point will be zero. At point P, the path difference between the rays coming from the slits S1 and S2 is S2P−S1P. Now, S1S2=d, EF=d, and S2F=D ∴ In ΔS2PF, S2P=[S2F2+PF2]12 S2P=[D2+(x+d2)2]12 = D⎡⎣1+(x+d2)2D2⎤⎦12 Similarly, in ΔS1PE S1P=D⎡⎣1+(x−d2)2D2⎤⎦12 S2P−S1P=D⎡⎣1+(x+d2)2D2⎤⎦12−D⎡⎣1+(x−d2)2D2⎤⎦12 On expanding it binomially, S2P−S1P=12D[4x×d2]=xdD For bright fringes (constructive interference), the path difference is an integral multiple of wavelengths, i.e., path difference is nλ. ∴ , nλ=xdD x=nλDd, where n=0, 1, 2, 3, 4,........... For n=0, x0=0 n=1, x1=λDd n=2, x2=2λDd n=3, x3=3λDd ⋮ n=n, xn=nλDd The separation between the centres of two consecutive bright interference fringes is the width of a dark fringe. ∴ β1=xn−xn−1=λDd Similarly, for dark fringes, xn=(2n−1)λD2d For n=1, x1=λD2d For n=2, x2=3λD2d The separation between the centres of two consecutive dark interference fringes is the width of a bright fringe. ∴ β2=xn−xn−1=λDd ∴ β1=β2 All the bright and dark fringes are of equal width as β1=β2 (b) Let ω, a and I represent the slit width, amplitude and intensity respectively. IminImax=(a1−a2)2(a1+a2)2=925 (a1−a2)(a1+a2)=35 (a1−a2)+(a1+a2)(a1+a2)−(a1−a2)=3+55−3 2a12a2=82 or a1a2=41 and ω1ω2=(a1)2(a2)2=161

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