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Discuss the diffraction at a grating and obtain the condition for the n^(th) maximum. |
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Answer» Solution :A plane transmission grating is reperented by AB. Let a plane wavefront of monochromatic light with wavelenght `LAMBDA` be incident normally on the grating. As the slits SIZE is comparable to that of wavelenght, the incident light diffracts at the grating. A diffraction patteern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens. Let US consider a point P at an angle `theta` with the normal drwan from the center of the gratng t the screen. The path difference `delta` between the diffracted waves from one pair of corresponding points is, `delta = (a + b) sin theta` This path difference is the same for any pair of corresponding points. The points P will be bright, when `delta = m lambda` where m = 0,1,2,3 Combining the above two equaitons, we get, `(a + b) sin theta = m lambda` Here, m is called order of diffraction. Condition for zero maximum, m = 0 For `(a + b) sin theta = 0`. the position, `theta = 0, sin 0 = 0 and m = 0`. This is called zero order diffraction or central maximum. Conditionfor second order maximum, m = 1, If `(a + b) sin thata_(1) = lambda`, the diffracted light MEET at an angle `theta_(1)` to the incidet direction and the first order maximum is obtained. Condition for second order maximum, m = 2 Similarly, `(a + b) sin theta_(2) = 2 lambda` forms the second order maximum at the angula positio `theta_(2)`. Condition for higher order maximum On EITHER side of central maxima different higher orders of diffraction maxima are formed at different angular position. It we take, `N = (1)/(a + b)` Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is specified on the grating itselt. Now, the equation becomes, `(1)/(N)sintheta=mlambda(or)sintheta=Nmlambda` 
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