Two coherent monochromatic point sources S1 and S2 of wavelength lambda = 600 nm are placed symmetri

Q: Why is the path difference at $P_1$ zero?

Because $P_1$ lies on the perpendicular bisector of the line joining the two sources $S_1$ and $S_2$. Any point on this bisector is equidistant from both sources, making the path difference exactly zero.

Q: Why do we use $d \sin\theta$ for the path difference?

When the radius of the circle is much larger than the separation $d$ between the sources, the rays reaching a point on the circle are approximately parallel. Dropping a perpendicular from one source to the other ray gives a path difference of $d \sin\theta$.

Q: What does it mean for the order of the fringe to be maximum at $P_2$?

The order of a fringe depends on the path difference. Since the maximum possible path difference in this setup is the distance $d$ between the sources (which occurs at $P_2$), the fringe order $n = d/\lambda$ is the highest possible value here.

Q: Why does the angular separation increase towards $P_2$?

Mathematically, the angular separation is inversely proportional to $\cos\theta$. As we move towards $P_2$, $\theta$ approaches $90^\circ$, making $\cos\theta$ smaller and the separation larger. Physically, the path difference changes more slowly as we move along the circle near $P_2$, so we have to travel a larger angle to see the next fringe.

Q: Could there be a dark spot at $P_2$?

A dark spot occurs when the path difference is a half-integer multiple of the wavelength. Since our path difference at $P_2$ is exactly 3000 times the wavelength (an integer), it forms a bright spot, not a dark one.

Question

Two coherent monochromatic point sources $S_1$ and $S_2$ of wavelength $\lambda = 600 	ext{ nm}$ are placed symmetrically on either side of the centre of the circle as shown. The sources are separated by a distance $d = 1.8 	ext{ mm}$. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is $\Delta	heta$. Which of the following options is/are correct?

Options:

(A) The angular separation between two consecutive bright spots decreases as we move from $P_1$ to $P_2$ along the first quadrant

(B) A dark spot will be formed at the point $P_2$

(C) The total number of fringes produced between $P_1$ and $P_2$ in the first quadrant is close to 3000

(D) At $P_2$ the order of the fringe will be maximum

Sigma Percentile · Accepted Answer

Interference patterns are usually studied on a flat screen, but what happens when the screen wraps around the sources in a perfect circle? This problem takes the classic Young's Double-Slit Experiment and bends it into a fascinating geometric puzzle. Let's break it down step by step and uncover the physics hidden along the circumference.

### Analyzing the Setup

To understand the interference pattern, our primary tool is the **path difference** ($\Delta x$) between the waves arriving from the two coherent sources, $S_1$ and $S_2$.

Let's first look at the extreme points on our circular screen. At point $P_1$, located on the vertical axis, the distances from $S_1$ and $S_2$ are exactly equal. This perfect symmetry means the path difference is zero ($\Delta x_{P_1} = 0$). Consequently, $P_1$ acts as the **central bright fringe** (or the zero-order maxima).

Now, imagine moving along the circle to point $P_2$ on the horizontal axis. Here, the path difference reaches its absolute maximum. The extra distance the wave from $S_1$ has to travel compared to the wave from $S_2$ is simply the physical separation between the sources themselves. Thus, at $P_2$, the path difference is $\Delta x_{P_2} = d = 1.8 	ext{ mm}$.

### The Master Equation

To determine what kind of fringe forms at $P_2$, we need to find its order, $n$. The order is the ratio of the path difference to the wavelength:
$$n = \frac{\Delta x_{P_2}}{\lambda}$$

Let's substitute the given values. The distance $d$ is $1.8 	imes 10^{-3} 	ext{ m}$, and the wavelength $\lambda$ is $600 	imes 10^{-9} 	ext{ m}$.
$$n = \frac{1.8 	imes 10^{-3}}{600 	imes 10^{-9}} = 3000$$

Since $n$ is a perfect integer, we can confidently say that a **bright fringe** is formed at $P_2$. This immediately tells us that option (b), which suggests a dark spot, is incorrect.

Furthermore, because the path difference at $P_2$ is the maximum possible in this setup, the order $n = 3000$ is the **maximum order** of the fringe. This confirms that option (d) is correct.

What about the fringes in between? As we move from $P_1$ (order 0) to $P_2$ (order 3000) in the first quadrant, we encounter every integer order from 1 to 2999. This means the total number of fringes produced between $P_1$ and $P_2$ is exactly 3000, making option (c) correct as well.

### Angular Separation

The most intriguing part of this problem is understanding how the fringes are spaced out. For a general point $P$ on the circle at an angle $	heta$ from the vertical axis, the path difference is approximately:
$$\Delta x = d \sin	heta$$

For a bright fringe of order $n$, we equate this to $n\lambda$:
$$d \sin	heta_n = n\lambda \implies \sin	heta_n = \frac{n\lambda}{d}$$

This tells us that for consecutive bright fringes, the difference in their sines is constant:
$$\sin	heta_n - \sin	heta_{n-1} = \frac{\lambda}{d}$$

To find the actual angular separation $\Delta	heta$, we can differentiate the equation $\sin	heta = \frac{n\lambda}{d}$. The derivative of $\sin	heta$ is $\cos	heta$, giving us:
$$\cos	heta \cdot \Delta	heta \approx \frac{\lambda}{d}$$
$$\Delta	heta \approx \frac{\lambda}{d \cos	heta}$$

This is our master equation for angular separation. Let's analyze its behavior. As we move from $P_1$ to $P_2$, the angle $	heta$ increases from $0^\circ$ to $90^\circ$. In this domain, the value of $\cos	heta$ steadily **decreases** towards zero.

Because $\cos	heta$ is in the denominator, a decreasing denominator means the overall fraction **increases**. Therefore, the angular separation $\Delta	heta$ must increase as we approach $P_2$. The fringes spread further apart! This proves that option (a) is incorrect.

### Conclusion

By carefully tracking the path difference and applying a touch of calculus, we've mapped out the entire interference pattern on a circular screen. The beauty of physics lies in how the mathematical equations perfectly mirror the physical reality—showing us exactly why the fringes bunch up near the top and spread out near the sides.

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Visualized Solution

$Visualizing the Setup$

$Path Difference at P_{1}$

$Path Difference at P_{2}$

$Fringe Order at P_{2}$

$Evaluating Options (b), (c), (d)$

$Path Difference at General Point P$

$Angular Separation$

$Evaluating Option (a)$

The Sigma Insight: Interference and Young's Double-Slit Experiment

Analyzing the Setup

The Master Equation

Angular Separation

Conclusion

Similar Questions

The light waves from two coherent sources have same intensity $I_{1} = I_{2} = I_{0}$ . In interference pattern the intensity of light at minima is zero. What will be the intensity of light at maxima ?

In a Young's double slit experiment, the slits are placed $0.320 mm$ apart. Light of wavelength $λ = 500 nm$ is incident on the slits. The total number of bright fringes that are observed in the angular range $- 3 0^{\circ} \leq θ \leq 3 0^{\circ}$ is

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Difference at P1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Difference at P2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fringe Order at P2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Options (b), (c), (d)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Difference at General Point P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Separation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Option (a)