Question

Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

• A. (i) and (iii) only
• B. (iii) and (iv) only
• C. (i), (iii) and (iv) only
• D. All of them

Hint:

For interference, remember the keyword: \[ \boxed{\text{Same frequency + Constant phase difference}} \] Different frequencies do not produce sustained interference.

Answer 1

The Correct Option is C

To solve the given question about interference between waves, we need to consider the basic requirement for interference. For two or more waves to interfere, they must be coherent, which means they must have a constant phase difference and the same frequency.

1. For waves \(y_1 = A_1\sin\omega t\) and \(y_2 = A_2\sin 2\omega t\), the frequencies are \(\omega\) and \(2\omega\) respectively. Since they have different frequencies, they cannot interfere.
2. For wave \(y_3 = A_3\cos\omega t\), if we convert this to a sine function, we get: y_3 = A_3\sin\left(\omega t + \frac{\pi}{2}\right). This wave has the same frequency \(\omega\) (as \(y_1\)), but with a phase difference.
3. For wave \(y_4 = A_4\sin\left(\omega t + \frac{\pi}{3}\right)\), the frequency is \(\omega\). It can interfere with \(y_1\) or \(y_3\) because they all share the same frequency, with different phase differences.

Now, we analyze pairs for possible interference:

• (i) and (iii): Both have the same frequency \(\omega\). Therefore, interference is possible.
• (iii) and (iv): Both have the same frequency \(\omega\). Therefore, interference is possible.
• (i), (iii), and (iv): All are of the same frequency \(\omega\), making interference possible between all three.

Waves (i) and (iv) can also interfere because they share the same frequency, and any phase difference only affects the interference pattern, not the possibility. Hence, the answer is (i), (iii), and (iv) only.