Single-Slit Diffraction: Minima and Central Maximum Width

1. The Geometry of Diffraction: Locating the Minima

Let's begin by setting up the problem. Light of a single wavelength $\lambda$ passes through a narrow slit of width $a$ and falls on a screen placed at a distance $D$ from the slit.

According to Huygens' principle, we can imagine the slit as being filled with an infinite number of secondary wavelet sources. The pattern on the screen is the result of all these wavelets interfering with each other.

Calculating the position of the bright fringes (maxima) is complex, but finding the dark fringes (minima) is more straightforward. A minimum occurs when there is perfect destructive interference.

Condition for the First Minimum

Consider the first dark fringe, which occurs at an angle $\theta$. For perfect cancellation, we can use a clever pairing argument. Imagine dividing the slit into two halves. If the wavelet from the very top of theslit travels exactly half a wavelength ($\lambda /2$) farther than the wavelet from the center of the slit, they will cancel out.

This condition holds for every pair of wavelets separated by a distance of $a/2$. The wavelet just below the top edge will cancel with the wavelet just below the center, and so on, for the entire slit. Therefore, the entire slit produces no light at that angle.

For this to happen, the path difference between the wavelet at the top edge and the wavelet at the bottom edge must be exactly one full wavelength, $\lambda$. From the geometry in the diagram, this path difference is $a\sin \theta$.

So, for the first minimum:
$a\sin \theta =\lambda$

General Condition for Minima

We can extend this logic. To find the second minimum, we can divide the slit into four equal parts. The first and second parts will cancel each other out, and the third and fourth parts will also cancel each other out, if the path difference between the top and bottom edges of the slit is $2\lambda$.

In general, a dark fringe (minimum) will be formed at an angle ${\theta }_n$ whenever the path difference is an integer multiple of the wavelength:

$a\sin {\theta }_n=n\lambda ,\text{for }n=\pm 1,\pm 2,\pm 3,\dots$

Important Note: The value $n=0$ is excluded because $\theta =0$ corresponds to the center of the pattern, where all wavelets arrive in phase, creating the bright central maximum, not a minimum.

2. Width of the Central Maximum

Now we can use the condition for the minima to define the size of the central maximum. The central maximum is the intensely bright region flanked by the first minimum on either side ($n=1$ and $n=-1$).

Angular Width

The first minimum occurs at an angle ${\theta }_1$ given by $a\sin {\theta }_1=(1)\lambda$.
$\sin {\theta }_1=\frac{\lambda }{a}$
The central maximum extends from $-{\theta }_1$ to $+{\theta }_1$. Therefore, its total angular width is $2{\theta }_1$.

In most practical scenarios, the angle $\theta$ is very small. This allows us to use the small angle approximation: $\sin \theta \approx \theta$ (where $\theta$ is in radians).

Applying this approximation, the angular position of the first minimum is:
${\theta }_1\approx \frac{\lambda }{a}$
The total angular width of the central maximum is therefore:
$\text{Angular Width}=2{\theta }_1\approx \frac{2\lambda }{a}$

Linear Width

The linear width, $W$, is the physical width of the central maximum on the screen. If $y_1$ is the distance from the center of the screen to the first minimum, then from trigonometry, $\tan {\theta }_1=y_1/D$.

Using the small angle approximation again ($\tan \theta \approx \theta$), we get:
${\theta }_1\approx \frac{y_1}{D}$
We can now equate our two expressions for ${\theta }_1$:
$\frac{y_1}{D}\approx \frac{\lambda }{a}$
Solving for $y_1$, we find the position of the first minimum:
$y_1\approx \frac{\lambda D}{a}$
The total linear width $W$ is twice this distance, as it spans from $-y_1$ to $+y_1$.
$\text{Linear Width}=W=2y_1\approx \frac{2\lambda D}{a}$

These formulas are fundamental for solving NEET problems. Notice the inverse relationship: a narrower slit (smaller $a$) produces a wider central maximum (larger $W$).

The derivation and formulas are summarized well in the following resource.

3. Solving NEET-Style Problems

Your goal is to become proficient in applying these formulas. Let's work through some examples to see how they are used in practice.

After watching the video, you can see the standard approach:

1. Identify the given variables ($\lambda ,a,D,W$).
2. Determine what you need to find.
3. Choose the appropriate formula: a sin θ = nλ for angles, or W = 2λD/a for linear width (assuming small angles).
4. Substitute the values and solve, being careful with units (e.g., converting nm and mm to meters).

Conclusion

In this lesson, you have moved from a qualitative to a quantitative understanding of single-slit diffraction. By focusing on the conditions for destructive interference, we derived the key formulas for the positions of the minima and the width of the central maximum—essential tools for your exam preparation.

Key Takeaways:

• Condition for Minima: Dark fringes occur at angles ${\theta }_n$ where $a\sin {\theta }_n=n\lambda$, for integers $n=\pm 1,\pm 2,\dots$.
• Angular Width of Central Maximum: The angular spread between the first minima on either side is approximately $2\lambda /a$.
• Linear Width of Central Maximum: The width on a screen at distance $D$ is approximately $W=2\lambda D/a$.
• Inverse Relationship: A narrower slit produces a wider diffraction pattern.

Preview of the Next Lesson:
The fact that light spreads out due to diffraction has a profound consequence: it limits our ability to see fine details. A telescope or microscope cannot form a perfect point image of a star; it forms a diffraction pattern instead. If two stars are very close, their diffraction patterns may overlap so much that we can't tell them apart. In our next lesson, we will explore this limitation by discussing the concept of resolving power and applying the Rayleigh criterion to optical instruments.