Problem 46 - Physics

Question

In a Young's double-slit experiment, the seventh dark fringe is located \(0.025 \mathrm{~m}\) to the side of the central bright fringe on a flat screen, which is \(1.1 \mathrm{~m}\) away from the slits. The separation between the slits is \(1.4 \times 10^{-4} \mathrm{~m}\). What is the wavelength of the light being used?

Step-by-Step Solution

Answer

The wavelength of the light is approximately 424 nm.

Step 1: Understand the Problem

We need to find the wavelength of light in a Young's double-slit experiment given the position of the seventh dark fringe, the distance from the slits to the screen, and the separation between the slits.

Step 2: Use the Dark Fringe Condition

The condition for dark fringes in a double-slit experiment is given by \( d \sin \theta = (m+\frac{1}{2}) \lambda \), where \( d \) is the slit separation, \( \theta \) is the angle of the fringe, \( m \) is the order of the dark fringe, and \( \lambda \) is the wavelength.

Step 3: Relate \\theta and Position on the Screen

Since the screen distance is much larger than the separation between fringes, use \( \tan \theta \approx \sin \theta \approx \frac{x_m}{L} \), where \( x_m \) is the fringe position from the central maximum, and \( L \) is the distance from the slits to the screen.

Step 4: Substitute Known Values

Plug the given values into the equation: \( d \left(\frac{x_m}{L}\right) = (m+\frac{1}{2}) \lambda \), where \(d = 1.4 \times 10^{-4} \mathrm{~m} \), \( x_m = 0.025 \mathrm{~m} \), \( L = 1.1 \mathrm{~m} \), and \( m = 7 \).

Step 5: Solve for the Wavelength

Rearrange the equation to solve for \( \lambda \): \( \lambda = \frac{d \cdot x_m}{L \cdot (m+\frac{1}{2})} \). Substituting in the values gives \[ \lambda = \frac{1.4 \times 10^{-4} \times 0.025}{1.1 \times 7.5} \approx 4.24 \times 10^{-7} \mathrm{~m} = 424 \mathrm{~nm}. \]

Key Concepts

Dark Fringe

In Young's double-slit experiment, dark fringes occur where destructive interference happens. Destructive interference means that the waves from the two slits are out of phase and cancel each other out. This cancellation leads to dark spots on the screen. These spots or fringes are observed as a series of parallel dark lines or bands. Each dark fringe corresponds to a specific order number, denoted by \( m \), where \( m = 0, 1, 2, \ldots \).

In our example, the seventh dark fringe, where \( m = 7 \), is relevant. The equation used to describe the condition for dark fringes in these scenarios is:
\[ d \sin \theta = \left(m + \frac{1}{2}\right) \lambda \]
Here, \( d \) is the slit separation, \( \theta \) is the angle of the dark fringe, and \( \lambda \) is the wavelength. Understanding this principle helps us identify where dark fringes will form and how they relate to the wavelength of the light used.

Wavelength Calculation

In the problem, we calculated the light wavelength using the position of the dark fringe. This requires rearranging the formula for dark fringes to solve for the wavelength \( \lambda \). Here's a step-by-step breakdown:

1. The condition for a dark fringe is given by:
\[ d \sin \theta = \left(m + \frac{1}{2}\right) \lambda \]
2. Due to small angles, we approximate \( \sin \theta \approx \tan \theta \approx \frac{x_m}{L} \), where \( x_m \) is the dark fringe distance from the central maximum, and \( L \) is the distance from the slits to the screen.
3. Substituting the given values, \( m = 7 \), \( d = 1.4 \times 10^{-4} \: \mathrm{m} \), \( x_m = 0.025 \: \mathrm{m} \), and \( L = 1.1 \: \mathrm{m} \), results in the equation:
\[ \lambda = \frac{d \cdot x_m}{L \cdot \left(m + \frac{1}{2}\right)} \]
4. Plugging in the numbers gives us:
\[ \lambda = \frac{1.4 \times 10^{-4} \times 0.025}{1.1 \times 7.5} \approx 4.24 \times 10^{-7} \: \mathrm{m} \approx 424 \: \mathrm{nm} \]

This wavelength falls within the visible spectrum of light, illustrating the direct connection between physical measurements in the experiment and the characteristics of the light used.

Interference Pattern

The interference pattern in Young's double-slit experiment showcases a beautiful display of alternating bright and dark bands on a screen. This pattern results from the overlapping of light waves exiting the two slits. Let's simplify how this pattern forms:

- **Constructive Interference:** Occurs where the waves reinforce each other, producing bright fringes.
- **Destructive Interference:** Causes waves to cancel each other out, creating dark fringes.

This pattern is valuable beyond its visual appeal. It acts as a fingerprint for determining the wavelength, because the distances between the fringes are directly related to the wavelength of the illuminating light.

The spacing of the peaks and troughs in the interference pattern depends on a few critical factors, including slit separation and wavelength. Thus, analyzing this pattern allows us to backtrack and find the wavelength when the other elements of the setup are known. It highlights the wave nature of light in a tangible and discoverable way.

Slit Separation

Slit separation, denoted by \( d \), is a crucial component in Young's double-slit experiment. It is the distance between the two slits through which the light is waved. This measurement impacts the interference pattern seen on the screen.

Here’s how slit separation affects the experiment:

- **Influence on Fringe Spacing:** The separation \( d \) inversely determines the distance between bright or dark fringes. Larger \( d \) values result in closer fringes, whereas smaller \( d \) values spread them apart.

- **Role in Wavelength Calculation:** Calculating the wavelength from the observed pattern involves \( d \), as seen in the equation \( d \sin \theta = \left(m + \frac{1}{2}\right) \lambda \). Thus, knowing \( d \) is essential for accurate measurements.

The initial setup of the experiment demands precise knowledge of this distance to predict the resulting interference pattern accurately. This precision highlights the intricate relationship between physical setup and observable phenomena in wave physics.