A coherent parallel beam of microwaves of wavelength λ = 0.5 mm falls on a Young's double slit appar

Q: Why can't we use the standard formula $y = \frac{(2n-1)\lambda D}{2d}$ for the position of minima?

That formula relies on the small-angle approximation ($\sin \theta \approx \tan \theta \approx \theta$), which is only valid when $\lambda \ll d$. Here, $\lambda = 0.5$ mm and $d = 1.0$ mm, so $\theta$ is quite large. We must calculate $\sin \theta$ exactly and then use geometry to find $y = D \tan \theta$.

Q: How did we know there are only two minima on each side in part (a)?

The physical constraint is that $\sin \theta$ cannot be greater than 1. When we set up the equation $\sin \theta = \frac{2n-1}{4} \le 1$, it restricts $n$ to be less than or equal to 2.5. Since $n$ must be an integer, only $n=1$ and $n=2$ are physically possible.

Q: What does the oblique incidence actually do to the interference pattern?

It gives one of the waves a 'head start'. The wave reaching the lower slit travels an extra distance before passing through. This shifts the entire interference pattern upwards on the screen, moving the central maximum to the angle of incidence.

Q: Why are the answers for part (b) the same as part (a)?

It's a beautiful mathematical coincidence in this specific setup! The initial path difference introduced by the $30^\circ$ angle is exactly $1\lambda$. Because the shift is an exact integer multiple of the wavelength, the new positions of the minima perfectly align with the old positions of the minima, even though the central maximum has moved.

Q: Is the central maximum always at the center of the screen?

Not always! The central maximum is defined as the point where the *net* path difference is zero. If the light hits the slits at an angle, the central maximum will shift to compensate for the initial path difference, as we saw in part (b).

Question

A coherent parallel beam of microwaves of wavelength $\lambda = 0.5 	ext{ mm}$ falls on a Young's double slit apparatus. The separation between the slits is $1.0 	ext{ mm}$. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of $1.0 	ext{ m}$ from it as shown in the figure.
(a) If the incident beam falls normally on the double slit apparatus, find the $y$-coordinates of all the interference minima on the screen.
(b) If the incident beam makes an angle of $30^\circ$ with the $x$-axis (as in the dotted arrow shown in figure), find the $y$-coordinates of the first minima on either side of the central maximum.

Sigma Percentile · Accepted Answer

### Riding the Microwaves: A Twist on Young's Double Slit Experiment

Welcome to a fascinating variation of the classic Young's Double Slit Experiment (YDSE). Instead of using visible light, we are exploring the interference pattern created by microwaves!

This seemingly small change has a massive impact on the mathematics of the problem. Because the wavelength of the microwaves ($\lambda = 0.5 	ext{ mm}$) is comparable to the slit separation ($d = 1.0 	ext{ mm}$), the standard small-angle approximation ($y = \frac{n\lambda D}{d}$) completely breaks down. We must rely on exact trigonometric relationships to find the true positions of the interference fringes.

### Part A: Normal Incidence and the Boundary Condition

When the microwaves fall normally on the slits, the path difference at a point $P$ on the screen, located at an angle $	heta$, is given by the standard geometric relation:

$$\Delta x = d \sin 	heta$$

For destructive interference (minima), this path difference must be an odd multiple of half the wavelength:

$$d \sin 	heta = (2n - 1)\frac{\lambda}{2}$$

Substituting the given values, we get:

$$\sin 	heta = \frac{(2n - 1)\lambda}{2d} = \frac{(2n - 1)(0.5)}{2(1.0)} = \frac{2n - 1}{4}$$

Here is where the physics imposes a strict mathematical boundary. The sine of any angle can never exceed $1$. This physical constraint limits the number of fringes we can observe:

$$\frac{2n - 1}{4} \le 1 \implies 2n - 1 \le 4 \implies n \le 2.5$$

Since $n$ must be an integer, the only valid values are $n = 1$ and $n = 2$. This means there are exactly two minima on either side of the central maximum!

Let's calculate their exact positions using $y = D 	an 	heta$:

**For the first minimum ($n=1$):**
$$\sin 	heta_1 = \frac{1}{4} \implies 	an 	heta_1 = \frac{1}{\sqrt{15}}$$
$$y_1 = 1.0 	imes \frac{1}{\sqrt{15}} \approx 0.26 	ext{ m}$$

**For the second minimum ($n=2$):**
$$\sin 	heta_2 = \frac{3}{4} \implies 	an 	heta_2 = \frac{3}{\sqrt{7}}$$
$$y_2 = 1.0 	imes \frac{3}{\sqrt{7}} \approx 1.13 	ext{ m}$$

Because the pattern is symmetric, the $y$-coordinates of all the interference minima are **$\pm 0.26 	ext{ m}$ and $\pm 1.13 	ext{ m}$**.

### Part B: Oblique Incidence and the Shifted Pattern

Now, let's tilt the incident beam by an angle $\alpha = 30^\circ$. This introduces an initial path difference before the waves even reach the slits. The wave reaching the lower slit travels an extra distance:

$$\Delta x_{initial} = d \sin \alpha = 1.0 	imes \sin 30^\circ = 0.5 	ext{ mm}$$

Notice something beautiful? This initial path difference is exactly equal to one full wavelength ($\lambda = 0.5 	ext{ mm}$)!

The net path difference at the screen is now the difference between the path difference created by the slits and the initial path difference:

$$\Delta x_{net} = d \sin 	heta - d \sin \alpha$$

The central maximum occurs where the net path difference is zero. This happens when $	heta = \alpha = 30^\circ$. The entire pattern has shifted upwards, placing the new central maximum at $y_0 = D 	an 30^\circ \approx 0.58 	ext{ m}$.

### The Beautiful Symmetry of the Minima

We need to find the first minima on either side of this shifted central maximum. For these points, the net path difference must be $\pm \frac{\lambda}{2}$:

$$d \sin 	heta - \lambda = \pm \frac{\lambda}{2}$$

Solving for $d \sin 	heta$, we get:

$$d \sin 	heta = \lambda \pm \frac{\lambda}{2} = \frac{3\lambda}{2} 	ext{ or } \frac{\lambda}{2}$$

Substituting the values back in, we find:

$$\sin 	heta = \frac{3}{4} \quad 	ext{and} \quad \sin 	heta = \frac{1}{4}$$

Look closely at these sine values! They are exactly the same values we found in Part A. Because the initial path difference was an exact integer multiple of the wavelength, the new positions of the minima perfectly align with the old positions of the minima.

Therefore, the $y$-coordinates of the first minima on either side of the central maximum are exactly **$0.26 	ext{ m}$ and $1.13 	ext{ m}$**.

Visualized Solution

$YDSE with Microwaves$

$Normal Incidence: Path Difference$

$Finding Valid Minima$

$Position of First Minima (n = 1)$

$Position of Second Minima (n = 2)$

$Oblique Incidence: Initial Path Difference$

$Shift of Central Maximum$

$First Minima for Oblique Incidence$

$Positions of Minima$

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

Riding the Microwaves

Part A

Part B

The Beautiful Symmetry of the Minima

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position of First Minima (n=1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position of Second Minima (n=2)

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Minima for Oblique Incidence

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