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Light with wavelength lambda falls normally on a long rectangular slit of width b. Find the angular distribution of the intensity of light in the case of Fraunhofer diffraction, as well as the angular position of minima. |
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Answer» Solution :If a plane wave is incident normally from the left on a slit of width `b` and the diffracted wave is observed at a large distance, the resulting pattern is called Fraunhofer diffraction. The condition for this is `b^(2) lt lt l lambda` where `l` is the distance between the slit and the screen. In paractice light MAY be focussed on the screen with the HELP of a lens(or a telescope). Consider an element of the slit which is an infinite strip of width `dx`. We use the formula of problem`5.103` with the following modifications. The factor `(1)/(r )` characteristic of spherical waves will be comiited. The factor `K (varphi)` wil also be dropped if we confine overselves to not too large `barphi`. In the DIRECTION defined by the angle `varphi` the extra path difference of the wave emitted from the element at `x` relative to the wave emitted from the centre is `Delta =- sin varphi` Thus the amplitude of the wave is given by `a UNDERSET(-b//2)overset(+b//2)int e^(ik sin varphi) dx = (e^(i(1)/(2)kb sin varphi)-e^(-i(1)/(2)kb sin varphi))//ik sin varphi` `= sin((pib)/(lambda)sin varphi)/((pib)/(lambda)sin varphi)` Thus `I = I_(0)(sin^(2)alpha)/(alpha^(2))` where `alpha = (pib)/(lambda) sin varphi` and `I_(0)` is a constant Minima are observed for `sin alpha = 0` but `alpha ~~ 0` Thus we find minima at angles given by `b sin varphi = k lambda, k = +-1,+-1, +-2,+-3,........` 
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