Question:
In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of that of Q by \(90^{\circ}\). A, B and C are three distinct points of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio :
In the figure below, P and Q are two equally intense coherent sources emitting radiation of wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of that of Q by \(90^{\circ}\). A, B and C are three distinct points of observation, each equidistant from the midpoint of PQ. The intensities of radiation at A, B, C will be in the ratio :
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Find the net phase difference at each point: source phase difference plus the phase due to path difference \(\frac{2\pi}{\lambda}\Delta x\), then use \(I = 4I_0\cos^2(\phi/2)\).
Updated On: Nov 5, 2026
- \(4 : 1 : 0\)
- \(2 : 1 : 0\)
- \(0 : 1 : 2\)
- \(0 : 1 : 4\)
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The Correct Option is B
Solution and Explanation
From the figure the order along the line is C, P, Q, A, and B lies on the perpendicular bisector of PQ.
Wave number \(k = \frac{2\pi}{20}\), so a path difference of 5 m corresponds to a phase \(k\cdot 5 = \frac{\pi}{2}\).
Phase difference at a point: \(\phi = (\phi_P - kr_P) - (\phi_Q - kr_Q) = \frac{\pi}{2} - k(r_P - r_Q)\).
Point A (on the Q side): \(r_P - r_Q = 5\) m, so \(\phi = \frac{\pi}{2} - \frac{\pi}{2} = 0\), \(I_A = 4I_0\).
Point B (on the bisector): \(r_P = r_Q\), so \(\phi = \frac{\pi}{2}\), \(I_B = 4I_0\cos^2\frac{\pi}{4} = 2I_0\).
Point C (on the P side): \(r_P - r_Q = -5\) m, so \(\phi = \frac{\pi}{2}+\frac{\pi}{2} = \pi\), \(I_C = 0\).
\(I_A : I_B : I_C = 4 : 2 : 0 = 2 : 1 : 0\).
Answer: (2) \(2 : 1 : 0\)
Wave number \(k = \frac{2\pi}{20}\), so a path difference of 5 m corresponds to a phase \(k\cdot 5 = \frac{\pi}{2}\).
Phase difference at a point: \(\phi = (\phi_P - kr_P) - (\phi_Q - kr_Q) = \frac{\pi}{2} - k(r_P - r_Q)\).
Point A (on the Q side): \(r_P - r_Q = 5\) m, so \(\phi = \frac{\pi}{2} - \frac{\pi}{2} = 0\), \(I_A = 4I_0\).
Point B (on the bisector): \(r_P = r_Q\), so \(\phi = \frac{\pi}{2}\), \(I_B = 4I_0\cos^2\frac{\pi}{4} = 2I_0\).
Point C (on the P side): \(r_P - r_Q = -5\) m, so \(\phi = \frac{\pi}{2}+\frac{\pi}{2} = \pi\), \(I_C = 0\).
\(I_A : I_B : I_C = 4 : 2 : 0 = 2 : 1 : 0\).
Answer: (2) \(2 : 1 : 0\)
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