Question

An interference is observed due to two coherent sources $A$ and $B$, having zero phase difference and separated by $4\lambda$, along the y-axis as shown in the figure (where $\lambda$ is the wavelength of the source). A detector $D$ is moved on the positive $x$-axis. The number of points on the $x$-axis, excluding the point $x=0$ and $x=\infty$, at which a maxima will be observed, are,

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Given:
Phase difference $\Delta \varphi =0$
Wavelength of the light =$\lambda$
Distance between $AB=4\lambda$

Let the path difference between the $AD$ and $BD$ is $\Delta x$.
A maxima will be observed when $\Delta x=n\lambda$ ($n=$integer)



From the diagram, for any point on positive side of x-axis, we can write,
$AD-BD=n\lambda$
$\Rightarrow AD=n\lambda +BD$
$\Rightarrow \sqrt{(4\lambda {)}^2+B{D}^2}=n\lambda +BD$

$\Rightarrow 16{\lambda }^2+B{D}^2=(n\lambda +BD{)}^2$
$\Rightarrow 16{\lambda }^2+B{D}^2=(n\lambda {)}^2+B{D}^2+2n\lambda \times BD$
$\Rightarrow {\lambda }^2(16-n^2)-2n\lambda BD=0$
$\Rightarrow \lambda (16-n^2)-2nBD=0$
$\Rightarrow BD=\frac{(16-n^2)\lambda }{2n}$

Now substituting the values of $n$,

For $n=0,BD=\infty$

For $n=1,BD=\frac{15\lambda }{2}=7.5\lambda$

For $n=2,BD=\frac{12\lambda }{4}=3\lambda$

For $n=3,BD=\frac{7\lambda }{6}$

For $n=4,BD=0$

Hence excluding the point $x=0$ and $x=\infty$ we get one maxima, which occurs for $n=2$

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

$$\begin{array}{c}\text{Why this question?}\\ \text{Tip : The resultant wave will attain maximum when path difference is equal to integral multiple of wavelength.}\end{array}$$