What is Central Maxima Width? | Simple Explanation for Class 12

S7-SA4-0301

What is the Width of Central Maxima in Single Slit Diffraction?

Grade Level:

Class 12

AI/ML, Physics, Biotechnology, FinTech, EVs, Space Technology, Climate Science, Blockchain, Medicine, Engineering, Law, Economics

Definition

What is it?

The width of the central maxima in single-slit diffraction is the size of the brightest and widest band of light formed on a screen when light passes through a narrow opening. It is the distance between the first dark fringes (minima) on either side of the center.

Simple Example

Quick Example

Imagine shining a torch through a very tiny gap, like a thin slit in a cardboard piece. Instead of a sharp line of light, you see a wider, bright band in the middle, fading out to dimmer lines on the sides. The width of that main, brightest band is what we're talking about.

Worked Example

Step-by-Step

Let's calculate the width of the central maxima.

Given:
- Wavelength of light (lambda) = 600 nm = 600 x 10^-9 meters
- Slit width (a) = 0.2 mm = 0.2 x 10^-3 meters
- Distance to screen (D) = 1 meter

Step 1: Understand the formula for the position of the first minimum. For single-slit diffraction, the angular position of the first minimum (theta) is given by sin(theta) = lambda / a. For small angles, sin(theta) is approximately theta.
---
Step 2: Calculate the angular position of the first minimum. theta = lambda / a = (600 x 10^-9 m) / (0.2 x 10^-3 m) = 3 x 10^-3 radians.
---
Step 3: The linear distance (y) of the first minimum from the center on the screen is given by y = D * tan(theta). For small angles, tan(theta) is approximately theta. So, y = D * theta.
---
Step 4: Calculate the distance of the first minimum from the center. y = 1 m * (3 x 10^-3 radians) = 3 x 10^-3 meters = 3 mm.
---
Step 5: The central maxima extends from the first minimum on one side to the first minimum on the other side. Therefore, the total width (W) of the central maxima is 2 * y.
---
Step 6: Calculate the width of the central maxima. W = 2 * (3 mm) = 6 mm.

Answer: The width of the central maxima is 6 mm.

Why It Matters

Understanding diffraction is key in designing high-tech devices like optical sensors used in AI/ML for image recognition, or in biotechnology for microscopes to see tiny cells. Engineers use this concept to build better cameras and display screens for your mobile phones and TVs, making images clearer.

Common Mistakes

MISTAKE: Confusing the width of the central maxima with the distance to the first minimum. | CORRECTION: The width of the central maxima is TWICE the distance from the center to the first minimum (2y).

MISTAKE: Using the formula for double-slit interference instead of single-slit diffraction. | CORRECTION: Remember that single-slit diffraction has a unique formula for minima positions (a sin(theta) = n lambda) and a much wider central maxima.

MISTAKE: Forgetting to convert units (like mm to meters, or nm to meters) before calculation. | CORRECTION: Always ensure all values are in consistent SI units (meters, seconds, kilograms) before plugging them into the formula.

Practice Questions

Try It Yourself

QUESTION: If the wavelength of light is 500 nm, the slit width is 0.1 mm, and the screen is 0.5 meters away, what is the distance of the first dark fringe from the center? | ANSWER: 2.5 mm

QUESTION: A single slit of width 0.4 mm is illuminated by light of wavelength 480 nm. If the central maxima has a width of 4.8 mm on a screen, what is the distance between the slit and the screen? | ANSWER: 2 meters

QUESTION: How would the width of the central maxima change if the wavelength of light used is doubled, while keeping the slit width and screen distance constant? Justify your answer. | ANSWER: The width of the central maxima would double. This is because the width is directly proportional to the wavelength (W = 2 * D * lambda / a).

MCQ

Quick Quiz

Which of the following changes would DECREASE the width of the central maxima in single-slit diffraction?

Increasing the wavelength of light

Decreasing the slit width

Increasing the distance to the screen

Increasing the slit width

The Correct Answer Is:

The width of the central maxima (W) is given by W = 2 * D * lambda / a. To decrease W, you need to either decrease D, decrease lambda, or increase a (slit width).

Real World Connection

In the Real World

This concept is crucial in fields like optical communication and medical imaging. For instance, in an eye clinic, doctors use instruments that rely on light diffraction to examine your retina. Also, advanced camera lenses in your smartphone use principles of diffraction to focus light and capture clear photos, even in low light.

Key Vocabulary

Key Terms

Diffraction: The bending of waves (like light) as they pass around an obstacle or through an opening. | Central Maxima: The brightest and widest band of light formed at the center of a diffraction pattern. | Slit Width: The actual width of the narrow opening through which light passes. | Wavelength: The distance between two consecutive crests or troughs of a wave.

What's Next

What to Learn Next

Great job understanding the central maxima! Next, you should explore 'Double Slit Interference'. This will help you see how multiple slits create different patterns, building on your knowledge of how light waves interact.