Question

Light waves from two coherent sources superimposes at a point. The waves at this point can be expressed as $y_1=a\sin \left({10}^{15}\pi t\right)$ and $y_2=2a\sin ({10}^{15}\pi t+\varphi )$. Here all quantities are in SI units.
Find the resultant amplitude $\left(\text{in m}\right)$, and frequency $\left(\text{in Hz}\right)$, of the resultant wave, if phase difference $\varphi$ is equal to $\pi /3$.

• A. $a\sqrt{5},5\times {10}^{14}$
• B. $a\sqrt{7},5\times {10}^{14}$
• C. $a\sqrt{5},{10}^{15}$
• D. $a\sqrt{7},{10}^{15}$

Answer 1

The correct option is B $a\sqrt{7},5\times {10}^{14}$

Resultant amplitude can be obtained from the relation,
$A^2=A_1^2+A_2^2+2A_1{A}_2\cos \varphi$

Where, $A_1$ and $A_2$ are the amplitudes of the interfering waves, $\varphi$ is the phase difference at the given point
and, $A$ is the resultant amplitude.

Here $A_1=a$, $A_2=2a$ and $\varphi =\frac{\pi }{3}$

$\Rightarrow A^2=a^2+4a^2+2\times a\times \left(2a\right)\cos \frac{\pi }{3}$
$\Rightarrow A^2=7a^2$
Or, $A=\sqrt{7}a$

Interference results due to superposition of waves of same frequency and frequency of the resultant wave also has the same value in the given problem.
Here, for each wave,
$\omega ={10}^{15}\pi$
$\Rightarrow 2\pi f={10}^{15}\pi$
$\therefore f=5\times {10}^{14}\text{Hz}$
Hence, the resultant frequency is $f=5\times {10}^{14}\text{Hz}$

Hence, $\left(B\right)$ is the correct answer.