Physics - Wave Optics Question with Solution | TestHub

Let source B be at $(0,0)$ and source A be at $(0,4\lambda )$.

The path difference $\Delta p$ at a point $D(x,0)$ on the positive X-axis is given by:

$\Delta p=AD-BD=\sqrt{x^2+(4\lambda {)}^2}-x$

Let us check the boundary values of the path difference:
At $x=0$, $\Delta p=\sqrt{0+16{\lambda }^2}-0=4\lambda$.
As $x\to \infty$, $\Delta p={\lim }_{x\to \infty }x{\left(1+\frac{16{\lambda }^2}{x^2}\right)}^{1/2}-x=0$.

For constructive interference (maxima), the path difference must be an integer multiple of the wavelength:

$\Delta p=n\lambda$

Since the points $x=0$ and $x\to \infty$ are explicitly excluded, the path difference must strictly lie in the range:

$0<\Delta p<4\lambda$

Therefore, the possible values for $\Delta p$ are $1\lambda ,2\lambda ,$ and $3\lambda$.

This corresponds to $n=1,2,3$.

Thus, there are exactly $3$ points on the positive X-axis where maxima will be observed.

Answer: $3$