Q124 / 150 · MHT CET 2026 (20 April Shift 2) · Physics

Two waves are represented as
$y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda}\right)$ and
$y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \dfrac{\pi}{6}\right)$
The path difference between two waves is

  • A
    $\dfrac{\lambda}{5}$
  • B
    $\dfrac{\lambda}{4}$
  • Correct answer
    C
    $\dfrac{\lambda}{3}$
  • D
    $\dfrac{\lambda}{2}$

Solution

The given equations of the waves are:

$y_1 = a_1\sin\left(\omega t - \dfrac{2\pi x}{\lambda}\right)$

$y_2 = a_2\cos\left(\omega t - \dfrac{2\pi x}{\lambda} + \dfrac{\pi}{6}\right)$

To find the phase difference, we express $y_2$ in terms of the sine function:

$y_2 = a_2\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \dfrac{\pi}{6} + \dfrac{\pi}{2}\right)$

$y_2 = a_2\sin\left(\omega t - \dfrac{2\pi x}{\lambda} + \dfrac{2\pi}{3}\right)$

The phase difference $\Delta \phi$ between the two waves is:

$\Delta \phi = \dfrac{2\pi}{3}$

The relation between path difference $\Delta x$ and phase difference $\Delta \phi$ is given by:

$\Delta x = \dfrac{\lambda}{2\pi} \Delta \phi$

Substituting the value of $\Delta \phi$:

$\Delta x = \dfrac{\lambda}{2\pi} \times \dfrac{2\pi}{3} = \dfrac{\lambda}{3}$

Answer: $\dfrac{\lambda}{3}$

Hence, the correct answer is option: 'C'.

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