Monochromatic light of 605 nm falls on a single slit of width 0.095 mm. The slit is located 85 cm from a screen. How far from the center is the first dark band? O 0.5 mm O 5.4 mm O 540 mm O 5,400 mm
Monochromatic light of 605 nm falls on a single slit of width 0.095 mm. The slit is located 85 cm from a screen. How far from the center is the first dark band? O 0.5 mm O 5.4 mm O 540 mm O 5,400 mm
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 and Interference: Single Slit Experiment**
You may use a calculator.
**Problem:**
Monochromatic light of 605 nm falls on a single slit of width 0.095 mm. The slit is located 85 cm from a screen. How far from the center is the first dark band?
**Options:**
- ⭕ 0.5 mm
- ⭕ 5.4 mm
- ⭕ 540 mm
- ⭕ 5,400 mm
**Explanation:**
In this problem, you'll apply the single-slit diffraction formula to determine the position of the first dark band. For a single slit, the condition for the first dark band (minimum) on the diffraction pattern is given by:
\[ a \cdot \sin(\theta) = m \cdot \lambda \]
Where:
- \( a \) is the width of the slit (0.095 mm).
- \( \lambda \) is the wavelength of the light (605 nm).
- \( m \) is the order of the minimum (for the first dark band, \( m = 1 \)).
- \( \theta \) is the angle relative to the central maximum.
The position of the dark band on the screen, \( y \), can be related to the angle \( \theta \) and the distance \( L \) from the slit to the screen using trigonometry:
\[ y = L \cdot \tan(\theta) \]
For small angles, \( \tan(\theta) \approx \sin(\theta) \), so:
\[ y = L \cdot \sin(\theta) \]
Substitute \( \sin(\theta) = \frac{m \cdot \lambda}{a} \):
\[ y = L \cdot \frac{m \cdot \lambda}{a} \]
Given:
- \( a = 0.095 \) mm \( = 0.095 \times 10^{-3} \) m
- \( \lambda = 605 \) nm \( = 605 \times 10^{-9} \) m
- \( L = 85 \) cm \( = 0.85 \) m
- \( m = 1 \)
Calculate the position \( y \):
\[ y = 0.85 \cdot \frac{1 \cdot 605 \times 10^{-9}}{0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8dc949f-7f1d-4c1b-9c4e-dfcc47554f56%2Fbbce5161-9be8-4e8e-9380-0706fbe6e7e7%2Fcvrqhe_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Diffraction and Interference: Single Slit Experiment**
You may use a calculator.
**Problem:**
Monochromatic light of 605 nm falls on a single slit of width 0.095 mm. The slit is located 85 cm from a screen. How far from the center is the first dark band?
**Options:**
- ⭕ 0.5 mm
- ⭕ 5.4 mm
- ⭕ 540 mm
- ⭕ 5,400 mm
**Explanation:**
In this problem, you'll apply the single-slit diffraction formula to determine the position of the first dark band. For a single slit, the condition for the first dark band (minimum) on the diffraction pattern is given by:
\[ a \cdot \sin(\theta) = m \cdot \lambda \]
Where:
- \( a \) is the width of the slit (0.095 mm).
- \( \lambda \) is the wavelength of the light (605 nm).
- \( m \) is the order of the minimum (for the first dark band, \( m = 1 \)).
- \( \theta \) is the angle relative to the central maximum.
The position of the dark band on the screen, \( y \), can be related to the angle \( \theta \) and the distance \( L \) from the slit to the screen using trigonometry:
\[ y = L \cdot \tan(\theta) \]
For small angles, \( \tan(\theta) \approx \sin(\theta) \), so:
\[ y = L \cdot \sin(\theta) \]
Substitute \( \sin(\theta) = \frac{m \cdot \lambda}{a} \):
\[ y = L \cdot \frac{m \cdot \lambda}{a} \]
Given:
- \( a = 0.095 \) mm \( = 0.095 \times 10^{-3} \) m
- \( \lambda = 605 \) nm \( = 605 \times 10^{-9} \) m
- \( L = 85 \) cm \( = 0.85 \) m
- \( m = 1 \)
Calculate the position \( y \):
\[ y = 0.85 \cdot \frac{1 \cdot 605 \times 10^{-9}}{0.
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