Light shines through a single slit whose width is 5.8 × 104 m. A diffraction pattern is formed on a flat screen located 4.0 m away. The distance between the middle of the central bright fringe and the first dark fringe is 3.9 mm. What is the wavelength of the light?
Light shines through a single slit whose width is 5.8 × 104 m. A diffraction pattern is formed on a flat screen located 4.0 m away. The distance between the middle of the central bright fringe and the first dark fringe is 3.9 mm. What is the wavelength of the light?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Diffraction Through a Single Slit**
When light passes through a single slit, a diffraction pattern is produced on a screen. This phenomenon occurs due to the wave nature of light.
### Problem Description
- **Slit Width**: \( 5.8 \times 10^{-4} \, \text{m} \)
- **Screen Distance**: 4.0 m away from the slit
- **Fringe Distance**: The distance between the middle of the central bright fringe and the first dark fringe is 3.9 mm
**Question**: What is the wavelength of the light?
### Formula
To find the wavelength (\(\lambda\)), we use the formula for the position of the first dark fringe in single slit diffraction:
\[
\lambda = \frac{a \cdot y_1}{L}
\]
Where:
- \( a \) = slit width
- \( y_1 \) = distance from the center to the first dark fringe
- \( L \) = distance from the slit to the screen
### Solution Steps
1. Convert \( y_1 \) from mm to meters:
\[
y_1 = 3.9 \, \text{mm} = 3.9 \times 10^{-3} \, \text{m}
\]
2. Substitute the given values into the formula:
\[
\lambda = \frac{(5.8 \times 10^{-4} \, \text{m}) \cdot (3.9 \times 10^{-3} \, \text{m})}{4.0 \, \text{m}}
\]
3. Calculate \(\lambda\).
This setup demonstrates the principle of single slit diffraction and allows calculation of the wavelength, an important characteristic of light.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f0a09bd-ef71-43b6-abcd-653755da1dfb%2F4444c8c6-4ff1-41e2-8a60-59244ba3b742%2F67we1zn_processed.png&w=3840&q=75)
Transcribed Image Text:**Diffraction Through a Single Slit**
When light passes through a single slit, a diffraction pattern is produced on a screen. This phenomenon occurs due to the wave nature of light.
### Problem Description
- **Slit Width**: \( 5.8 \times 10^{-4} \, \text{m} \)
- **Screen Distance**: 4.0 m away from the slit
- **Fringe Distance**: The distance between the middle of the central bright fringe and the first dark fringe is 3.9 mm
**Question**: What is the wavelength of the light?
### Formula
To find the wavelength (\(\lambda\)), we use the formula for the position of the first dark fringe in single slit diffraction:
\[
\lambda = \frac{a \cdot y_1}{L}
\]
Where:
- \( a \) = slit width
- \( y_1 \) = distance from the center to the first dark fringe
- \( L \) = distance from the slit to the screen
### Solution Steps
1. Convert \( y_1 \) from mm to meters:
\[
y_1 = 3.9 \, \text{mm} = 3.9 \times 10^{-3} \, \text{m}
\]
2. Substitute the given values into the formula:
\[
\lambda = \frac{(5.8 \times 10^{-4} \, \text{m}) \cdot (3.9 \times 10^{-3} \, \text{m})}{4.0 \, \text{m}}
\]
3. Calculate \(\lambda\).
This setup demonstrates the principle of single slit diffraction and allows calculation of the wavelength, an important characteristic of light.
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