Chapter 22: Problem 39
Monochromatic light of unknown wavelength passes through two slits separated by \(0.12 \mathrm{~mm}\), producing an interference pattern on a screen \(1.55 \mathrm{~m}\) away. If the distance between the central maximum and the first dark fringe is \(3.40 \mathrm{~mm}\), what's the wavelength of the light?
Short Answer
Step by step solution
Understanding the Problem
Relationship Between Variables
Simplification Using Small Angle Approximation
Rearranging Formula
Substituting Known Values
Calculate Wavelength
Convert to Nanometers
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interference Pattern
This happens due to a key phenomenon: constructive and destructive interference.
- Constructive interference: When the peaks of two light waves meet, they add together, creating a bright spot.
- Destructive interference: When a peak and a trough meet, they cancel each other out, creating a dark spot.
Understanding this concept is crucial for analyzing patterns in experiments like the one we've discussed. The position of these bright and dark fringes tells us useful information, such as the wavelength of the light used.
Wavelength Calculation
In our exercise, we're given the distance between the slits, the distance from the slits to the screen, and the distance from the central maximum to the first dark fringe. With these pieces of information, we apply a specific formula derived from understanding the fringe pattern.
- The relationship used is: \[ \lambda = \frac{2 d x}{L} \]
- Where:
- \( d \) is the slit separation,
- \( x \) is the distance to the first dark fringe,
- \( L \) is the distance from the slits to the screen.
Substituting the known values directly provides us the wavelength of the light used in the experiment. Once you have this calculation, converting from meters to other units like nanometers can provide additional context for the light's characteristics.
Dark Fringe
Dark fringes occur because light from one slit cancels out light from the other slit at specific points. In our current exercise, we focus on the first dark fringe. This appears directly next to the central bright fringe, which is the primary bright spot aligned with the slits.
- Mathematically, for the first dark fringe in a double-slit experiment, it's accounted for by having \ - \( (m + 0.5) \lambda = d \sin \theta \) where \( m = 0 \).
Dark fringes are essential since they are used as reference points for calculations, such as finding the light's wavelength, as demonstrated in the given exercise. They show the incredible precision and predictability of wave interference!