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In deriving the single-slit diffraction pattern, it was states that the intensity is zero at angles (nlamda)/(a). Justify this by suitable dividing the slit to bring out the cancellation. |
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Answer» Solution :Let us consider a single-slit of width a. for diffraction pattern due this slit nth principal MINIMA will be obtained at an angle `theta` such that `asintheta=nlamda or sintheta=(nlamda)/(a)`, and for small angle `sintheta=theta=(nlamda)/(a)`. To justify this, let us consider that the given slit is divided into 2n smaller slits, each of EQUAL width `a.=(a)/(2n)`. then in the direction of diffraction angle `theta`, the path DIFFERENCE between wavelets from two adjacent slits will be `a.sintheta=((a)/(2n))sintheta=(a)/(2n)*(nlamda)/(a)=(lamda)/(2)`. it means that light from each odd NUMBERED slit is exactly nullified by the light from SUBSEQUENT even numbered slit and resultant intensity is zero. |
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