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A palne monochromatic ligth wave falls normally on a long rectangular slit behind which a screen is positioned at a distance b = 60 cm. First the width of the slit has adjucted so that in the middle of the diffraction pattern the lowest minimum was observed. After widening the slit by Deltah 0.70mm, the next minimum was obtained in the centre of the pattern. Find the wavelength of light. |
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Answer» Solution :If the aperture has width `h` them the parameters `(v, -v)` associted with `(h//2,-(h)/(2))` are given by `v = (h)/(2) sqrt((2)/(b lambda)) = h//sqrt(2b lambda)` the intensity of light at `O` on the screen is obtained as the square of the amplitude `A` of the wave at `O` which is `A_(~) const UNDERSET(-v)overset(v)int e^(-ipiu^(2)//2) du` Thus `I = 2I_(0) ((C(v))^(2) + (S'(v))^(2))` where `C(v)` and `S(v)` have been defind above and `I_(0)` is the intensity at `O` due to an infinitey wide `(v = oo)` aperture for then `I = 2I_(1) (((1)/(2))(2) + ((1)/(2))^(2)) = 2I_(0) xx (1)/(2) = I_(0)`. By DEFINITION `v` corresponds to the first minimum of the intensity. This means `v = v_(1) ~~90` when we increase `h` to `h + Deltah`, the corresponding `v_(2) = (h + Deltah)/(sqrt(2b lambda)` relates to the second minimum of intensity. From the cornu's spiral `v_(2) ~~2.75` Thus `DELTA h = sqrt(2b lambda) (v_(2) - v_(1)) = 0.85 sqrt(2b lambda)` or `lambda = ((Deltah)/(0.85))^(2) (1)/(2b) = ((0.70)/(0.85))^(2) (1)/(2xx0.6)mu m = 0.565 mu m` 
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