Problem 9 - Physics

Question

A sheet that is made of plastic \((n=1.60)\) covers one slit of a double slit (see the drawing). When the double slit is illuminated by monochromatic light \(\left(\lambda_{\text {vacuum }}=586 \mathrm{nm}\right),\) the center of the screen appears dark rather than bright. What is the minimum thickness of the plastic?

Step-by-Step Solution

Answer

The minimum thickness of the plastic is 488.33 nm.

Step 1: Understand the problem situation

We're given a double-slit experiment where a plastic sheet with refractive index \(n = 1.60\) covers one of the slits. Monochromatic light of wavelength \( \lambda = 586 \text{ nm} \) is used, and the center of the screen appears dark, indicating destructive interference. We need to determine the minimum thickness of the plastic sheet so that the path difference leads to this result.

Step 2: Conceptualize destructive interference

For the center of the screen to appear dark, there must be a phase difference equivalent to half a wavelength (a path difference of \(m + 0.5\) wavelengths, where \(m\) is an integer). In this case, the path difference is due to the optical thickness (\(t'(n-1)\)) of the plastic sheet.

Step 3: Calculate the optical path difference

The optical path difference needed for destructive interference is \((m + 0.5) \cdot \lambda\). For minimum thickness, let's consider \(m = 0\). Therefore, the optical path difference is \(0.5\cdot\lambda\). In terms of thickness and refractive index, this equation becomes \(t(n - 1) = 0.5 \cdot \lambda\).

Step 4: Solve for thickness \(t\)

Rearrange the equation from Step 3: \[ t(n - 1) = 0.5 \cdot \lambda \] Where \(\lambda\) is the wavelength in a vacuum. Substituting \(n = 1.60\) and \( \lambda = 586 \text{ nm} \): \[ t = \frac{0.5 \cdot 586 \text{ nm}}{1.60 - 1} = \frac{293.0 \text{ nm}}{0.60} \] Calculate: \[ t = 488.33 \text{ nm} \] This is the minimum thickness that will result in a dark center on the screen.

Key Concepts

Understanding Optical Path Difference

In a double-slit experiment, light waves travel different paths to reach a point on a screen. The difference in these paths is called the "optical path difference." This difference determines whether the light waves will interfere constructively or destructively.
An optical path difference is calculated by considering both the geometric path and the material through which the light passes. The formula can be expressed as:\[ \text{Optical Path Difference} = t(n-1) \]
Here, \( t \) is the material's thickness, and \( n \) is the refractive index. By controlling this difference, we can manipulate the interference pattern on a screen. To achieve destructive interference at the screen's center, the optical path difference must be half a wavelength. This calculation helps find the required thickness of the plastic covering one of the slits.

Exploring Destructive Interference

Destructive interference occurs when the crest of one wave meets the trough of another, canceling each other out. This results in a dark spot on the screen. For the experiment's central point to appear dark, the waves from both slits need to be out of phase by half a wavelength.

• When two waves are half a wavelength out of phase, they interfere destructively.
• This means the optical path difference is \( \frac{1}{2} \lambda \), where \( \lambda \) is the wavelength.

Destructive interference in a double-slit setup provides valuable insight into wave behavior. By knowing this concept, you can determine how to manipulate the interference pattern by introducing elements like a plastic sheet with a specific thickness.

Role of Refractive Index

The refractive index \( n \) indicates how much light slows down when passing through a material. This factor significantly influences optical path differences.

• A higher refractive index means light travels slower through the material.
• This also increases the effective optical path, contributing to interference effects.

Given a refractive index of 1.60, the plastic sheet causes light to slow down more compared to air, affecting the phase and resulting in destructive interference. Calculating the impact of refractive index helps in designing experiments and understanding how materials interact with light, ultimately leading to precise control over interference patterns.