Chapter 35: Problem 13
Two very narrow slits are spaced 1.80 \(\mu\)m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \(\lambda\) = 550 nm? (Hint: The angle \(\theta\) in Eq. (35.5) is \(not\) small.)
Short Answer
Step by step solution
Understand the Problem
Use the Condition for Dark Lines
Calculate the Angles \(\theta_1\) and \(\theta_2\)
Calculate Position of Dark Lines 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.
Destructive interference
In the context of our double-slit experiment, destructive interference results in dark lines on the screen where the waves from the two slits perfectly cancel each other out. This occurs when the path difference between the two waves is a half-multiple of the wavelength, specifically \[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]where:
- \(d\) is the distance between the slits,
- \(\lambda\) is the wavelength of the light,
- \(\theta\) is the angle of the dark fringe from the horizontal,
- \(m\) is an integer (0, 1, 2, ...).
Double-slit experiment
The resulting light waves overlap and interfere with each other, producing distinct patterns on a screen behind the slits. We see a series of bright and dark lines. Bright lines occur from constructive interference, while dark lines, where no light appears, result from destructive interference.
This experiment is not only foundational in understanding wave properties of light but also delves into the principles of superposition, where the resulting wave pattern is a sum of individual wave patterns. As in our problem, by knowing slit spacing and wavelength, interference patterns teach us about wave interactions on a detailed level.
Wavelength calculation
In calculating where dark lines form in interference patterns (as in our example), we use the formula specific to destructive interference:\[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]Here, \(m\) is the order of the dark fringe, and knowing the position \(\theta\) helps us locate these dark spots precisely on the screen.
To find where these lines appear on the screen, converting angles to linear position using \( y = L\tan\theta \), where \(L\) is the distance from the slits to the screen, is crucial. All these calculations revolve around understanding wavelength, showing its central role in predicting wave interference patterns effectively.