Single-Slit Diffraction and Related Problems

Single-Slit Diffraction and Related Problems

Single-slit diffraction is a fundamental phenomenon in wave optics, where light passes through a narrow slit and spreads out, creating a pattern of bright and dark regions on a screen. The diffraction pattern results from interference between light waves emerging from different parts of the slit.

1. Diffraction from a Single Slit

When monochromatic light passes through a single slit, it behaves as a set of point sources of secondary wavelets, according to Huygens' principle. These wavelets interfere with each other, leading to a characteristic diffraction pattern with a central bright fringe, flanked by alternating dark and bright fringes.

Key Features of the Single-Slit Diffraction Pattern:

• Central Maximum: The brightest and widest band, located directly in front of the slit.
• Secondary Maxima: Weaker bright fringes that occur between the dark fringes.
• Dark Fringes: Points where destructive interference occurs, creating completely dark spots.

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2. Condition for Dark Fringes (Destructive Interference)

The dark fringes in a single-slit diffraction pattern occur when the path difference between the light waves from the edges of the slit is an integer multiple of the wavelength. The condition for destructive interference (dark fringes) is given by:

$$a \sin(\theta) = m \lambda$$

Where:

• $a$ is the width of the slit,
• $\theta$ is the angle of diffraction,
• $\lambda$ is the wavelength of the light,
• $m$ is an integer (1, 2, 3, ...) representing the order of the minima (first, second, third, etc.).

For the central maximum (bright fringe), the light from all points in the slit arrives in phase, leading to constructive interference.

Angular Position of the Minima:

• The angular positions of the first minima (dark fringes) on either side of the central maximum can be found by substituting $m = 1, 2, 3, \dots$.

  For example:

  • The first dark fringe occurs when $m = 1$,
  • The second dark fringe occurs when $m = 2$, and so on.

Width of the Central Maximum:

The angular width of the central maximum (the region between the first minima on either side) is:

$$\Delta \theta = \frac{2\lambda}{a}$$

This is the angular separation between the first dark fringes on either side of the central maximum.

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3. Intensity Distribution in Single-Slit Diffraction

The intensity of light in the diffraction pattern varies across the screen, and the distribution of light intensity $I(\theta)$ as a function of the angle $\theta$ is given by:

$$I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2$$

Where:

• $I_0$ is the intensity at the central maximum,
• $\beta = \frac{\pi a}{\lambda} \sin(\theta)$,
• $\lambda$ is the wavelength of light,
• $a$ is the slit width,
• $\theta$ is the angle.

The central maximum is the brightest, and the intensity gradually decreases as you move away from the center. The first minima occurs at $\beta = \pi$, and subsequent minima occur at multiples of $\pi$.

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4. Single-Slit Diffraction Example Problems

Example 1: Angular Position of the First Minimum

Problem: A monochromatic light of wavelength $600 \, \text{nm}$ passes through a slit of width $0.1 \, \text{mm}$. What is the angle for the first dark fringe in the diffraction pattern if the screen is placed 5 meters away?

Solution:

• Given:
  • Wavelength of light: $\lambda = 600 \, \text{nm} = 6 \times 10^{-7} \, \text{m}$,
  • Slit width: $a = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m}$,
  • Order of the dark fringe: $m = 1$ (first dark fringe),
  • Distance to the screen: $L = 5 \, \text{m}$.

Using the condition for destructive interference:

$$a \sin(\theta) = m \lambda$$

For the first dark fringe ($m = 1$):

$$(1 \times 10^{-4}) \sin(\theta) = 6 \times 10^{-7}$$ $$\sin(\theta) = \frac{6 \times 10^{-7}}{1 \times 10^{-4}} = 6 \times 10^{-3}$$

For small angles, $\sin(\theta) \approx \theta$, so:

$$\theta \approx 6 \times 10^{-3} \, \text{radians}$$

To find the linear position $y$ on the screen for the first dark fringe, we use the approximation:

$$y = L \theta = 5 \times 6 \times 10^{-3} = 0.03 \, \text{m} = 3 \, \text{cm}$$

So, the first dark fringe occurs at an angle of approximately $6 \times 10^{-3}$ radians and is located $3 \, \text{cm}$ from the central maximum.

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Example 2: Width of the Central Maximum

Problem: A single-slit diffraction pattern is produced with light of wavelength $500 \, \text{nm}$ passing through a slit of width $0.25 \, \text{mm}$. What is the angular width of the central maximum?

Solution:

• Given:
  • Wavelength: $\lambda = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m}$,
  • Slit width: $a = 0.25 \, \text{mm} = 2.5 \times 10^{-4} \, \text{m}$.

The angular width of the central maximum is the angular separation between the first minima on both sides, which is:

$$\Delta \theta = \frac{2\lambda}{a}$$

Substituting the given values:

$$\Delta \theta = \frac{2 \times 5 \times 10^{-7}}{2.5 \times 10^{-4}} = 4 \times 10^{-3} \, \text{radians}$$

So, the angular width of the central maximum is $4 \times 10^{-3}$ radians, or $0.004 \, \text{radians}$.

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Example 3: Intensity at a Given Angle

Problem: A monochromatic light of wavelength $650 \, \text{nm}$ passes through a slit of width $0.2 \, \text{mm}$. What is the intensity at an angle $\theta = 30^\circ$ if the intensity at the central maximum is $I_0 = 100 \, \text{arbitrary units}$?

Solution:

• Given:
  • Wavelength: $\lambda = 650 \, \text{nm} = 6.5 \times 10^{-7} \, \text{m}$,
  • Slit width: $a = 0.2 \, \text{mm} = 2 \times 10^{-4} \, \text{m}$,
  • Angle: $\theta = 30^\circ$,
  • Intensity at the central maximum: $I_0 = 100 \, \text{arbitrary units}$.

First, calculate $\beta$:

$$\beta = \frac{\pi a}{\lambda} \sin(\theta)$$ $$\beta = \frac{\pi \times 2 \times 10^{-4}}{6.5 \times 10^{-7}} \times \sin(30^\circ) = \frac{\pi \times 2 \times 10^{-4}}{6.5 \times 10^{-7}} \times \frac{1}{2}$$ $$\beta = \frac{\pi \times 10^{-4}}{6.5 \times 10^{-7}} \approx 484.78$$

Now, using the intensity formula:

$$I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2$$

Since $\beta$ is large, $\sin(\beta) \approx \frac{1}{\beta}$, so:

$$I(\theta) \approx I_0 \left( \frac{1}{\beta} \right)^2 = 100 \times \left( \frac{1}{484.78} \right)^2$$ I(\theta) \approx 100 \times (4.12 \times 10^{-5})^2 = 100 \times 1.69 \