Problem 21 - Physics

Question

Light that has a wavelength of \(668 \mathrm{~nm}\) passes through a slit \(6.73 \times 10^{-6} \mathrm{~m}\) wide and falls on a screen that is \(1.85 \mathrm{~m}\) away. What is the distance on the screen from the center of the central bright fringe to the third dark fringe on either side?

Step-by-Step Solution

Answer

The distance to the third dark fringe is approximately 549 mm.

Step 1: Understanding Diffraction and the Formula

This problem involves single-slit diffraction. For a single slit, the position of dark fringes is given by the formula for minima: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \lambda \) is the wavelength, \( \theta \) is the angle, and \( m \) is the order of the dark fringe (1, 2, 3, ...). In this case, we need the third dark fringe, so \( m = 3 \).

Step 2: Calculation of the Angle \( \theta \) for the Third Dark Fringe

Using the formula \( a \sin \theta = m \lambda \), substitute \( a = 6.73 \times 10^{-6} \) m, \( \lambda = 668 \times 10^{-9} \) m, and \( m = 3 \). Calculate \( \sin \theta \):\[ \sin \theta = \frac{3 \times 668 \times 10^{-9}}{6.73 \times 10^{-6}} \approx 0.297 \].Since \( \sin \theta \) is relatively small, \( \theta \approx \sin \theta \) in radians.

Step 3: Determining the Distance from the Central Maximum

The distance to the fringe on the screen \( y \) can be found using the small angle approximation \( y = L \tan \theta \approx L \sin \theta \). Substitute \( L = 1.85 \) m and \( \sin \theta \approx 0.297 \):\[ y = 1.85 \times 0.297 = 0.54945 \text{ m} \approx 549 \text{ mm}. \]

Key Concepts

Understanding Diffraction Minima

In the realm of single-slit diffraction, the term "diffraction minima" is crucial. When light waves pass through a narrow slit, they spread out and create patterns of dark and bright areas called fringes. The dark fringes, where the light intensity is minimal, are known as diffraction minima. This occurs because of destructive interference, where certain parts of the wave overlap and cancel each other out.
To calculate the position of these minima on a screen, we use the formula:

• \(a \sin \theta = m \lambda\)

The variables include:

• \(a\): Slit width
• \(\theta\): Angle of diffraction
• \(m\): Order of the fringe
• \(\lambda\): Wavelength of light

For the third dark fringe \( m = 3 \). Remembering this formula is essential for understanding why light and dark patterns form when waves go through small openings.

Exploring the Wavelength of Light

The wavelength of light, denoted as \( \lambda \), is a major player in diffraction scenarios. It represents the distance between successive peaks of a wave. In this exercise, the light's wavelength is given as 668 nanometers (nm), which is a microscopic scale more commonly used in optics.
The wavelength directly influences where the diffraction minima occur, as evident in the formula \( a \sin \theta = m \lambda \). A longer wavelength leads to more spread out fringes, while a shorter wavelength results in fringes that are closely packed.
Understanding the wavelength is vital not only for solving problems but also for interpreting how light behaves when it interacts with different materials.

The Significance of Slit Width in Diffraction Patterns

Slit width, symbolized as \( a \), is a measure of how wide the opening is through which light passes. In our problem, the slit width is \(6.73 \times 10^{-6} \text{ m}\).
The width Plays a pivotal role in determining the diffraction pattern because it affects how significantly the waves spread out after passing through the slit.
A smaller slit width causes broader diffraction patterns, meaning the dark and bright fringes appear farther apart. Conversely, a wider slit results in closely spaced fringes.
The slit width appears in the diffraction minima formula, showing its direct impact on where the dark fringes will occur on a screen.

Calculating Distance to the Screen and its Role

Distance to the screen, often denoted as \( L \), is the distance from the slit to the observing surface where diffraction patterns are displayed. In this scenario, the screen is 1.85 meters away from the slit.
This distance is critical when calculating the actual position of the diffraction minima on the screen. Using the approximated angle \( \theta \) from earlier calculations, the physical distance \( y \) to the specific fringe can be determined using:

• \( y = L \sin \theta \)

A larger distance to the screen increases the distance between fringes, making the diffraction pattern appear larger and easier to observe. Conversely, a shorter distance results in a more compact pattern.