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InterviewSolution
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Discuss the special cases on first minimum in Fraunhofer diffraction. |
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Answer» Solution :Consider the condition for first minimum with (n = 1). a sin `theta = lamda` The first minimum has an angular spread of, `sin theta = lamda/a` Now, we have special cases to discuss on the above condition. (i) When `a lt lamda`, the diffraction is not possible, because `sin theta` can never be greater than 1. (ii) When `a gt= lamda` the diffraction is possible. (a) For `a = lamda, sin theta = 1 i.e., theta = 90^@` . That means the first minimum is at `90^@`. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to `90^@` . (b) For `a gt gtlamda, sin theta lt lt1` i.e, the first minimum will fall within the width of the slit itself. The diffraction will · not be noticed at all. (III) When `agtlamda` and also comparable, say `a = 2 lamda, sin theta =lamda/a = (lama)/(2 lamda) = 1/2` ` theta= 30^@` These are practical cases where diffraction could be observed effectively. (iv) The slit AB is divided into two half's AC and CB. Now the width of AC `(a/2)`. Different points on the slit whichare separated by the same width(here ` a/2` ) called corresponding points. (v)The path difference of light waves from different corresponding points meeting at POINT P and interfere destructively to make it first minimum. The path difference `delta` between waves from these corresponding points is, `delta = a/2 sin theta` (vi) The condition for P to be first minimum, ` a/2 sin theta = lamda/2` ` a sin theta = lamda ` (first minimum) . |
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