Question

The equation of two interfering light waves are
\(y_{1}=b\cos \omega t\) and \(y_{2}=b\sin (\omega t+\Phi )\).
If the light waves interfere destructively, then the possible value of \(\Phi\) are

• A. \(\frac{3\pi }{2},\frac{5\pi }{2},\frac{13\pi }{2},\ldots \ldots\)
• B. \(\frac{3\pi }{2},\frac{7\pi }{2},\frac{13\pi }{2},\ldots\)
• C. \(\frac{3\pi }{2},\frac{7\pi }{2},\frac{11\pi }{2},\ldots \ldots\)
• D. \(\frac{3\pi }{2},\frac{9\pi }{2},\frac{11\pi }{2},\ldots \ldots\)

Hint:

Write both waves as sine functions, find the phase difference, and set it equal to an odd multiple of pi.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Interference depends on the phase difference between the two waves.
Destructive interference needs a phase difference of an odd multiple of \(\pi\).

Step 2: Phase difference
Write \(y_{1}=b\cos\omega t=b\sin\left(\omega t+\dfrac{\pi}{2}\right)\).
The phase difference with \(y_{2}=b\sin(\omega t+\Phi)\) is \(\Phi-\dfrac{\pi}{2}\).

Step 3: Apply the condition
\(\Phi-\dfrac{\pi}{2}=(2n+1)\pi\), so \(\Phi=\dfrac{3\pi}{2},\dfrac{7\pi}{2},\dfrac{11\pi}{2},\ldots\)
Successive values differ by \(2\pi\).

Final Answer:
The possible values of \(\Phi\) are \(3\pi/2,\ 7\pi/2,\ 11\pi/2,\ldots\) \[ \boxed{\text{(C) }\dfrac{3\pi}{2},\dfrac{7\pi}{2},\dfrac{11\pi}{2},\ldots} \]