A) 20
B) 22
C) 24
D) 26
Correct Answer: A
Solution :
[a] Path difference at P is \[\Delta x=2\left( \frac{x}{2}\cos \theta \right)=x\cos \theta \] For intensity to be maximum, \[\Delta x=n\lambda \] (n=0, 1, 2, 3, ....) or\[x\cos \theta =n\lambda \] or \[x\,\cos \,\theta =\frac{n\lambda }{x}\ge 1\] \[\therefore \,\,\,n\ge \frac{x}{\lambda }\] Subsituting x= 51, we get \[n\ge 5\]or n= 1, 2, 3, 4, 5....... Therefore in all four quadrants there can be 20 maxima. There are more maxima at \[\theta =0{}^\circ \]and\[\theta =180{}^\circ \]. But n = 5 corresponds to \[\theta =90{}^\circ \] and \[\theta =270{}^\circ \] which are coming only twice while we have multuplies it four times. Therefore, total number of maxima are still 20, i.e., n = 1 to 4 in four quadrants (total 16) plus more at \[\theta =0{}^\circ \], \[90{}^\circ \], \[180{}^\circ \]and\[270{}^\circ \].You need to login to perform this action.
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