Question

What is the path difference between two waves $y_1=a_1\sin (\omega t-{\displaystyle \frac{2\pi x}{\lambda }})$ and $y_2=a_2\cos (\omega t-{\displaystyle \frac{2\pi x}{\lambda }}+\varphi )$ ?
$$\begin{gathered} \text{A}\text{. }{\displaystyle \frac{\lambda }{2\pi }}[\varphi +{\displaystyle \frac{\pi }{2}}] \\ \text{B}\text{.}{\displaystyle \frac{\lambda }{2\pi }}[\varphi ] \\ \text{C}\text{. }{\displaystyle \frac{\lambda }{2\pi }}[\varphi -{\displaystyle \frac{\pi }{2}}] \\ \text{D}\text{. }{\displaystyle \frac{2\pi }{\lambda }}[\varphi ] \end{gathered}$$

Answer 1

Hint: The two waves are given in different functions. So, convert both of them to either sine or cosine. Find the phase difference between them, then get the path difference.
Formula used:
$\cos \theta =\sin ({\displaystyle \frac{\pi }{2}}+\theta )$
$\mathrm{\Delta }x={\displaystyle \frac{\lambda }{2\pi }}\times \mathrm{\Delta }\varphi$ , where $\mathrm{\Delta }x,\mathrm{\Delta }\varphi$ are the path difference and phase difference respectively.
This shows that the phase difference and the path difference are directly proportional to each other.

Complete step-by-step solution -
Path difference is the difference in the distance traveled by two waves at the meeting point. It measures how much a wave is shifted from another.
The phase difference is simply the difference in the phase of the two traveling waves. The phase difference is an important property as it determines the nature of the interference pattern and diffraction pattern obtained.
If the path difference between two waves is an integral multiple $(n\lambda )$ of wavelength, we get constructive interference. When the path difference between two waves is an odd multiple of half wavelength$({\displaystyle \frac{(2n-1)\lambda }{2}})$ , we get destructive interference.
 $y_1=a_1\sin (\omega t-{\displaystyle \frac{2\pi x}{\lambda }})$
$y_2=a_2\cos (\omega t-{\displaystyle \frac{2\pi x}{\lambda }}+\varphi )$
Converting $y_2$ into sine function
$y_2=a_2\sin ({\displaystyle \frac{\pi }{2}}+(\omega t-{\displaystyle \frac{2\pi x}{\lambda }}+\varphi ))$
Subtract the angles of $y_{1}and{y}_2$
So, the phase difference is, ${\displaystyle \frac{\pi }{2}}+\varphi$
Now, substituting the value of phase difference in the formula
$\mathrm{\Delta }x={\displaystyle \frac{\lambda }{2\pi }}\times \mathrm{\Delta }\varphi$
We get the path difference as:
${\displaystyle \frac{\lambda }{2\pi }}\times [{\displaystyle \frac{\pi }{2}}+\varphi ]$
The correct option is (A).

Note: While converting sine to cosine or vice versa, take care of the sign. This is a simple formula based question, so remember the formula.