Miscellaneous: Practice Problem & Solution
Assertion (A): Distance between two coherent sources \(S_1\) and \(S_2\) is \(4\lambda\). A large circle is drawn around these sources with centre of circle lying on centre of \(S_1\) and \(S_2\). there are total 16 maxima on the circle. Reason (R): Total number of minima on this circle are less compare to total number of maximas.
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Miscellaneous. Understanding the underlying formula is key to arriving at the correct answer below:
Given distance \(d=4lambda\). Path difference \(\Delta x = dcos\theta = 4\lambdacos\theta\). The range for \(\Delta x\) is \(-4\lambda \le \Delta x \le 4\lambda\). For maxima, \(\Delta x = n\lambda\), so \(-4 \le n \le 4\). This gives 9 integer values for \(n\). \(n=pm 4\) correspond to \(cos\theta = pm 1\) (2 points). The other 7 values \(n=pm 3, pm 2, pm 1, 0\) correspond to \(|costheta|<1\) (giving \(2times 7 = 14\) points). Total maxima = \(2+14=16\). So (A) is true. For minima, \(\Delta x = (n+1/2)\lambda\), so \(-4.5 \le n \le 3.5\). This gives 8 integer values for \(n\) (from -4 to 3). For all these values, \(|(n+1/2)/4|<1\), so each gives 2 points. Total minima = \(8\times 2 = 16\). Thus, there are 16 maxima and 16 minima. Reason (R) is false.
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