Physics - Wave Optics Question with Solution | TestHub

Let $O$ be the origin on the screen. For any point $P$ on the screen at a distance $r$ from $O$, let the angle subtended by $P$ at the sources with respect to the central axis be $\theta$.

The path difference $\Delta x$ between the light waves reaching point $P$ from $S_1$ and $S_2$ is given by:
$\Delta x=S_{1}P-S_{2}P$

Since $D\gg d$, we can use the far-field approximation:
$\Delta x\approx d\cos \theta$

At point $O$, $r=0$, which means $\theta =0^{\circ }$. Thus, the path difference at $O$ is:
$\Delta x_O=d\cos (0^{\circ })=d$

As we move away from $O$ towards infinity on the screen, $\theta$ increases from $0$ to $\pi /2$, and the path difference $\Delta x$ continuously decreases from $d$ to $0$. Therefore, the path difference on the screen lies in the range $(0,d]$.

Evaluating each option:

For option (A):
If $d=\frac{7\lambda }{2}=3.5\lambda$, the path difference at $O$ is $\Delta x=3.5\lambda$. Since this is an odd multiple of $\lambda /2$, destructive interference occurs, making $O$ a minima. This statement is correct.

For option (B):
If $d=\lambda$, the path difference on the screen is in the range $(0,\lambda ]$. The condition for maxima is $\Delta x=n\lambda$. The only integer multiple of $\lambda$ in this range is for $n=1$, which gives $\Delta x=\lambda$ at point $O$. The maximum for $n=0$ ($\Delta x=0$) is formed at infinity, which is not observed on the screen. Thus, only one maximum is observed on the screen. This statement is correct.

For option (C):
If $d=4.8\lambda$, the path difference on the screen is in the range $(0,4.8\lambda ]$. The condition for minima is $\Delta x=(n-0.5)\lambda$. The possible values for minima in this range are $0.5\lambda ,1.5\lambda ,2.5\lambda ,3.5\lambda$, and $4.5\lambda$. These $5$ distinct path differences correspond to $5$ circular dark fringes (minima) on the screen. This statement is correct.

For option (D):
If $d=\lambda$, the path difference at $O$ is $\Delta x=\lambda$. Since this is an integer multiple of $\lambda$, constructive interference occurs, making the intensity at $O$ a maximum, not a minimum. This statement is incorrect.

Answer: If $d=\frac{7\lambda }{2},O$ will be a minima; If $d=\lambda$, only one maxima can be observed on the screen; If $d=4.8\lambda$, then total $5$ minima would be there on the screen