Question

There is a destructive interference between the two waves of wavelength $\lambda$ coming from two different paths at a point. To get the constructive interference at that point, the path of one wave is to be increased by,

• A. $\frac{\lambda }{4}$
  
• B. $\frac{\lambda }{2}$
  
• C. $\frac{3\lambda }{4}$
  
• D. $\lambda$
  

Answer 1

The correct option is B $\frac{\lambda }{2}$


Let the path difference between the two wave is $\Delta x$.
So, for destructive interference
$\Delta x=\frac{n\lambda }{2}...\left(1\right)$
Where, $n=1,3,5,...$

Let the path of one wave is increased by $y$ to get constructive interference.
The constructive interference is obtained when,
$\Delta x+y=n\lambda ...\left(2\right)$
Where $n=0,1,2,...$

Subtracting equation $\left(1\right)$ from $\left(2\right)$,
$\Rightarrow y=\frac{n\lambda }{2}$
Putting $n=1$, we get,
$y=\frac{\lambda }{2}$

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

$$\begin{array}{c}\text{Why this Question}\\ \text{Concept: The interference of two waves with identical wavelengths passing through a common point depends on their phase difference}\Delta \varphi \text{. Constructive interference is obtained when}\Delta \varphi \text{is integral multiple of}2\pi \text{and fully destructive interference occurs when}\Delta \varphi \text{is an odd multiple of}\pi \text{.}\end{array}$$