Chapter 36: Problem 21
An interference pattern is produced by light of wavelength 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 mm. (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit interference maxima? (b) Let the slits have width 0.320 mm. In terms of the intensity \(I_0\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?
Short Answer
Step by step solution
Understand the Problem
First-order Interference Maxima
Second-order Interference Maxima
Calculate Angular Positions
Consider Slit Width Influence on Intensity
Intensity at First-order Maximum
Intensity at Second-order Maximum
Evaluate Final Intensities
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Two-Slit Experiment
Explaining interference patterns is simple:
- Bright fringes appear when the peaks of light waves from the two slits align perfectly, leading to constructive interference. Here, the path difference between waves is a whole number multiple of wavelengths, denoted as \( m \lambda \).
- Dark fringes occur when the peak of one wave aligns with the trough of another, leading to destructive interference.
Wavelength
- In this exercise, the wavelength of the light used is 580 nm, or nanometers, which is typical for visible light.
- The role of wavelength in the interference pattern can be seen in how it relates to the path difference needed for constructive interference: \( m \lambda \).
Intensity
However, the actual intensity at any angular position is influenced by the slit width, introducing a single-slit diffraction component to the pattern.
- The intensity at an angular position \( \theta \) is given by the formula \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \).
- Here, \( \beta = \frac{\pi a \sin \theta}{\lambda} \), where \( a \) represents the slit width.
Angular Position
Using the formula for constructive interference, \( d \sin \theta = m \lambda \) (where \( d \) is the distance between slits, \( m \) is the order of maxima, and \( \lambda \) is the wavelength), we can determine these angular positions.
- For first-order maxima (\( m=1 \)), \( \theta_1 \) is calculated, and analogously, for second-order maxima (\( m=2 \)), \( \theta_2 \).
- These angles are small in most practical setups, leading to approximate values often expressed in degrees, such as \( 0.0627^\circ \) and \( 0.1254^\circ \).