Interference Fringes on Circular Screen with Two Point Sources

To determine which of the options are correct, let us analyze the path difference and interference pattern formed along the circumference of the circular screen.

1. Geometry and Path Difference

Let the line joining the two coherent sources $S_1$ and $S_2$ be along the horizontal axis (x-axis), with the center of the circle $O$ as the origin $(0,0)$.

The positions of the sources are given as: $S_1 = \left(-\frac{d}{2}, 0\right), \quad S_2 = \left(\frac{d}{2}, 0\right)$

where $d = 1.8 \text{ mm} = 1.8 \times 10^{-3} \text{ m}$ and the wavelength is $\lambda = 600 \text{ nm} = 6 \times 10^{-7} \text{ m}$.

For any point $P$ on the circular screen located at an angle $\phi$ relative to the line $S_1 S_2$ (i.e., relative to $OP_2$):

• At $P_2$, $\phi = 0^\circ$
• At $P_1$, $\phi = 90^\circ$

Assuming $R \gg d$, the path difference $\Delta x$ between the waves arriving at point $P$ from sources $S_1$ and $S_2$ is given by: $\Delta x = d \cos\phi$

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2. Analysis of Options

Option A & B: Fringe Order at $P_2$

At point $P_2$, the angle is $\phi = 0^\circ$. Therefore, the path difference is maximum: $\Delta x_{max} = d \cos(0^\circ) = d$

Let us calculate the order of fringe $m$ at $P_2$: $m = \frac{\Delta x_{max}}{\lambda} = \frac{d}{\lambda} = \frac{1.8 \times 10^{-3} \text{ m}}{6 \times 10^{-7} \text{ m}} = 3000$

• Since $m = 3000$ is an integer, a bright spot (central maximum of order 3000) is formed at point $P_2$. Thus, Option A is incorrect.
• The fringe order reaches its maximum possible value of $m_{max} = 3000$ at $P_2$. Thus, Option B is correct.

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Option C: Total Number of Fringes between $P_1$ and $P_2$

• At $P_1$ ($\phi = 90^\circ$), the path difference is $\Delta x = d \cos(90^\circ) = 0$, which corresponds to order $m = 0$ (a bright spot).
• At $P_2$ ($\phi = 0^\circ$), the path difference corresponds to order $m = 3000$.

The bright fringes formed between $P_1$ and $P_2$ in the first quadrant correspond to integers $m = 0, 1, 2, \dots, 3000$. Hence, the total number of bright fringes produced in the first quadrant is approximately 3000. Thus, Option C is correct.

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Option D: Variation of Angular Separation $\Delta\theta$

For a bright spot of order $m$: $d \cos\phi = m \lambda$

Differentiating both sides with respect to $m$: $-d \sin\phi \, d\phi = \lambda \, dm$

For two consecutive bright spots, $\Delta m = 1$. The magnitude of the angular separation $\Delta\phi$ is: $\Delta\phi = \frac{\lambda}{d \sin\phi}$

As we move from $P_1$ ($\phi = 90^\circ$) to $P_2$ ($\phi = 0^\circ$):

• At $P_1$, $\sin\phi = 1$, so $\Delta\phi$ is at its minimum value: $\Delta\phi_{min} = \frac{\lambda}{d}$.
• As $\phi$ decreases from $90^\circ$ to $0^\circ$, $\sin\phi$ decreases, causing the angular separation $\Delta\phi$ to increase.

Therefore, the angular separation between consecutive bright spots increases (not decreases) as we move from $P_1$ to $P_2$. Thus, Option D is incorrect.

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Conclusion

The correct options are B and C.