Chapter 26: Problem 11
\(\bullet\) Two slits spaced 0.450 \(\mathrm{mm}\) apart are placed 75.0 \(\mathrm{cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 \(\mathrm{nm} ?\)
Short Answer
Step by step solution
Understand the Problem
Determine the Condition for Dark Fringes
Calculate the Angles for the Second and Third Dark Lines
Calculate the Positions on the Screen
Find the Distance Between the Dark Lines
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Slit Experiment
The light from each slit acts as a single wave source. As these waves overlap on the screen, their peaks and troughs interact to produce an interference pattern. This pattern consists of alternating bright and dark bands, known as fringes, which can be explained by the path length difference between the light traveling through each slit. The Double Slit Experiment is critical for understanding wave interference phenomena.
Dark Fringes
The condition for a dark fringe is given by the equation:
- \[d \sin \theta = (m + 0.5) \lambda\]
- \(d\) is the distance between the slits,
- \(\theta\) is the angle relative to the central maximum,
- \(\lambda\) is the wavelength of the light, and
- \(m\) is the order of the dark fringe (an integer value).
Wavelength of Light
The wavelength determines the spacing of the interference fringes. In our given exercise, the wavelength of the light used is 500 nm (nanometers), which is a measure of how "long" the wave is. Shorter wavelengths, like blue light, produce more closely spaced fringes, while longer wavelengths, like red light, create wider fringes. Thus, understanding the wavelength is crucial for predicting the position of bright and dark fringes.
Small Angle Approximation
- \(\sin \theta \approx \tan \theta \approx \theta\)
- \(y \approx L \theta\)
In the provided exercise, using this approximation simplifies our task of locating the dark fringes, making it easier to handle without complex trigonometric functions. This is particularly helpful when the screen is far from the slits, ensuring angles remain small.