In an interference arrangement similar to Young's double-slit experiment, the slits S1 and S2 are il

Q: Why do we use the formula $I = I_0 \cos^2(\delta/2)$?

This is the standard result for the interference of two coherent waves of equal amplitude. The resultant amplitude is $2A \cos(\delta/2)$, and since intensity is proportional to the square of the amplitude, we get $I \propto 4A^2 \cos^2(\delta/2)$, which is $I_0 \cos^2(\delta/2)$.

Q: How did we get the path difference as $d \sin \theta$?

For a point very far away (Fraunhofer diffraction condition), the rays from the two slits are approximately parallel. If you drop a perpendicular from one slit to the ray from the other, the extra distance traveled by the second ray is $d \sin \theta$.

Q: Why is the wavelength so large (300 m)?

We are dealing with microwaves of frequency $10^6$ Hz (1 MHz), which is in the radio frequency range. Using $c = \nu \lambda$, the wavelength comes out to be 300 meters. This is why the slits are separated by 150 meters to observe interference!

Q: What happens if the sources are not synchronized?

If the sources are not synchronized, they would have a random, rapidly changing phase difference. This would average out the interference pattern, resulting in a uniform intensity of $I_0/2$ everywhere on the screen.

Q: Why is the intensity zero at $90^\circ$?

At $90^\circ$, the path difference is exactly $d$, which is 150 m. Since the wavelength is 300 m, this path difference is exactly $\lambda/2$. A path difference of half a wavelength corresponds to destructive interference, resulting in zero intensity.

Question

In an interference arrangement similar to Young's double-slit experiment, the slits $S_1$ and $S_2$ are illuminated with coherent microwave sources, each of frequency $10^6$ Hz. The sources are synchronized to have zero phase difference. The slits are separated by a distance $d = 150.0$ m. The intensity $I(	heta)$ is measured as a function of $	heta$, where $	heta$ is defined as shown.
If $I_0$ is the maximum intensity, then $I(	heta)$ for $0^\circ \le 	heta \le 90^\circ$ is given by

Options:

(A) $I(	heta) = I_0 / 2$ for $	heta = 30^\circ$

(B) $I(	heta) = I_0 / 4$ for $	heta = 90^\circ$

(C) $I(	heta) = I_0$ for $	heta = 0^\circ$

(D) $I(	heta)$ is constant for all values of $	heta$

Sigma Percentile · Accepted Answer

The phenomenon of interference is one of the most beautiful demonstrations of the wave nature of light—and in this case, microwaves! Let's dive into this classic Young's double-slit style problem and unravel the mystery of the intensity distribution.

### Analyzing the Setup

Imagine two slits, $S_1$ and $S_2$, acting as coherent sources of microwaves. They are separated by a massive distance of $d = 150 	ext{ m}$. The microwaves have a frequency of $
u = 10^6 	ext{ Hz}$.

Before we can determine the intensity at any angle $	heta$, we need to know the wavelength $\lambda$ of these microwaves. Using the fundamental wave equation $c = 
u \lambda$, we can find the wavelength:

$$ \lambda = \frac{c}{
u} = \frac{3 	imes 10^8 	ext{ m/s}}{10^6 	ext{ Hz}} = 300 	ext{ m} $$

This large wavelength is characteristic of radio-frequency waves, which explains why the slits are placed so far apart to observe the interference pattern!

### The Master Equation

The intensity $I(	heta)$ at any point in an interference pattern is governed by the phase difference $\delta$ between the waves arriving from the two slits. The master equation for intensity is:

$$ I(	heta) = I_0 \cos^2\left(\frac{\delta}{2}ight) $$

where $I_0$ is the maximum possible intensity.

The phase difference $\delta$ is directly proportional to the path difference $\Delta x$ between the two waves. For a point far away at an angle $	heta$, the path difference is simply $\Delta x = d \sin 	heta$. Therefore, the phase difference is:

$$ \delta = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} (d \sin 	heta) $$

Substituting our known values of $d = 150 	ext{ m}$ and $\lambda = 300 	ext{ m}$, we get a beautifully simplified expression for the phase difference:

$$ \delta = \frac{2\pi}{300} (150 \sin 	heta) = \pi \sin 	heta $$

### Testing the Angles

Now, let's test the given options one by one by plugging in the respective angles.

**For $	heta = 30^\circ$:**
The phase difference becomes:

$$ \delta = \pi \sin 30^\circ = \pi \left(\frac{1}{2}ight) = \frac{\pi}{2} $$

Plugging this into our intensity equation:

$$ I(30^\circ) = I_0 \cos^2\left(\frac{\pi/2}{2}ight) = I_0 \cos^2\left(\frac{\pi}{4}ight) $$

Since $\cos(\pi/4) = 1/\sqrt{2}$, squaring it gives $1/2$. Thus, $I(30^\circ) = I_0 / 2$. **Option (a) is correct!**

**For $	heta = 90^\circ$:**
The phase difference is:

$$ \delta = \pi \sin 90^\circ = \pi (1) = \pi $$

The intensity is:

$$ I(90^\circ) = I_0 \cos^2\left(\frac{\pi}{2}ight) = 0 $$

The intensity is zero, not $I_0 / 4$. **Option (b) is incorrect.**

**For $	heta = 0^\circ$:**
The phase difference is:

$$ \delta = \pi \sin 0^\circ = 0 $$

The intensity is:

$$ I(0^\circ) = I_0 \cos^2(0) = I_0 $$

This is the central maximum, where the intensity is indeed $I_0$. **Option (c) is correct!**

### Final Conclusion

Since the intensity $I(	heta)$ clearly depends on the angle $	heta$, it is not constant. Therefore, option (d) is incorrect. The correct options are **(a)** and **(c)**. This problem elegantly combines wave kinematics with the geometry of interference, proving that the principles of optics apply just as perfectly to microwaves as they do to visible light!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Intensity Formula$

$Phase Difference$

$Calculating Wavelength$

$Testing Option (a)$

$Intensity for Option (a)$

$Testing Option (b)$

$Testing Option (c)$

$Conclusion$

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

Analyzing the Setup

The Master Equation

Testing the Angles

Final Conclusion

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Select Answer:

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; } Intensity Formula

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

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

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

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

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

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

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