Question

Two coherent narrow slits emitting sound of wavelength $\lambda$, in the same phase are placed parallel to each other at a small separation of $2\lambda$. The sound is detected by moving a detector on the screen which is placed at a distance $d$ from the slit $S_1$ as shown in figure. Find the distance $y$ such that the intensity at $P$ is equal to intensity at point $O$ if $d>>\lambda$.

• A. $\sqrt{2}d$
• B. $\sqrt{3}d$
• C. $\sqrt{5}d$
• D. $2d$

Answer 1

The correct option is B $\sqrt{3}d$


From the figure, we can deduce that, path difference at ${}^{\prime }{O}^{\prime }$ is given by $S_{1}O-S_{2}O=2\lambda$.
So, constructive interference is obtained at $O$.

As $y\to \infty ,\Delta x\to 0$

Since path difference decreases from $2\lambda$ to $0$, next constructive interference will be obtained when $\Delta x=1\lambda$.

i.e. $|S_{1}P-S_{2}P|=\lambda$

From the diagram, we can write that,

$|\sqrt{d^2+y^2}-\sqrt{(d-2\lambda {)}^2+y^2}|=\lambda$

$\Rightarrow \sqrt{d^2+y^2}-\lambda =\sqrt{(d-2\lambda {)}^2+y^2}$

Squaring on both sides we get,

$d^2+y^2+{\lambda }^2-2\lambda \sqrt{d^2+y^2}=d^2+4{\lambda }^2-4\lambda d+y^2$

$\Rightarrow 3{\lambda }^2+2\lambda \sqrt{d^2+y^2}-4\lambda d=0$

$\Rightarrow 2\sqrt{d^2+y^2}=4d-3\lambda .........\left(1\right)$

As $d>>\lambda$

So, $2\sqrt{d^2+y^2}=4d$

$\Rightarrow d^2+y^2=4d^2\Rightarrow y=\sqrt{3}d$

Thus, option (b) is the correct answer.