Tomas uses light of wavelength 536 nm to illuminate a round 0.7-mm diameter hole. A screen is placed 6.7 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? (please give answer with one decimal place)
Tomas uses light of wavelength 536 nm to illuminate a round 0.7-mm diameter hole. A screen is placed 6.7 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? (please give answer with one decimal place)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter38: Diffraction Patterns And Polarization
Section: Chapter Questions
Problem 38.29P: The laser in a compact disc player must precisely follow the spiral track on CD, along which the...
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![### Diffraction of Light through a Circular Aperture
**Problem Statement:**
Tomas uses light of wavelength 536 nm to illuminate a round 0.7-mm diameter hole. A screen is placed 6.7 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? (Please give the answer with one decimal place)
**Analysis and Solution:**
This problem involves the phenomenon of diffraction, which occurs when a wave encounters an obstacle or a slit that is comparable in size to its wavelength. For a circular aperture, the diffraction pattern formed on the screen can be characterized by a central bright area (central maximum) surrounded by alternating dark and bright rings. The diameter of this central bright area can be calculated using the following formula for the first minimum (dark ring):
\[ \sin(\theta) = 1.22 \frac{\lambda}{D} \]
where:
- \(\theta\) is the angle from the central axis to the first minimum,
- \(\lambda\) is the wavelength of the light,
- \(D\) is the diameter of the aperture.
However, since the angle \(\theta\) is small, \(\sin(\theta) \approx \tan(\theta) \approx \theta\).
Using the small angle approximation, we can express the distance from the central bright spot to the first minimum on the screen as:
\[ y = L \tan(\theta) \approx L \theta = L \left(1.22 \frac{\lambda}{D}\right) \]
where \(L\) is the distance from the aperture to the screen.
1. **Calculate the angle \(\theta\):**
\[ \theta = 1.22 \frac{\lambda}{D} = 1.22 \frac{536 \times 10^{-9} \, \text{m}}{0.7 \times 10^{-3} \, \text{m}} \approx 9.34 \times 10^{-4} \, \text{radians} \]
2. **Calculate the distance y to the first minimum on one side:**
\[ y = 6.7 \, \text{m} \times 9.34 \times 10^{-4} \approx 6.2 \times 10^{-3} \, \text{m} = 6.2 \, \text{mm}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F620ec3ab-ba57-4203-af2d-0e5f479038c2%2Fe31cf4d1-495b-4fb4-935b-cb873c7692c6%2Fqse9cv5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Diffraction of Light through a Circular Aperture
**Problem Statement:**
Tomas uses light of wavelength 536 nm to illuminate a round 0.7-mm diameter hole. A screen is placed 6.7 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? (Please give the answer with one decimal place)
**Analysis and Solution:**
This problem involves the phenomenon of diffraction, which occurs when a wave encounters an obstacle or a slit that is comparable in size to its wavelength. For a circular aperture, the diffraction pattern formed on the screen can be characterized by a central bright area (central maximum) surrounded by alternating dark and bright rings. The diameter of this central bright area can be calculated using the following formula for the first minimum (dark ring):
\[ \sin(\theta) = 1.22 \frac{\lambda}{D} \]
where:
- \(\theta\) is the angle from the central axis to the first minimum,
- \(\lambda\) is the wavelength of the light,
- \(D\) is the diameter of the aperture.
However, since the angle \(\theta\) is small, \(\sin(\theta) \approx \tan(\theta) \approx \theta\).
Using the small angle approximation, we can express the distance from the central bright spot to the first minimum on the screen as:
\[ y = L \tan(\theta) \approx L \theta = L \left(1.22 \frac{\lambda}{D}\right) \]
where \(L\) is the distance from the aperture to the screen.
1. **Calculate the angle \(\theta\):**
\[ \theta = 1.22 \frac{\lambda}{D} = 1.22 \frac{536 \times 10^{-9} \, \text{m}}{0.7 \times 10^{-3} \, \text{m}} \approx 9.34 \times 10^{-4} \, \text{radians} \]
2. **Calculate the distance y to the first minimum on one side:**
\[ y = 6.7 \, \text{m} \times 9.34 \times 10^{-4} \approx 6.2 \times 10^{-3} \, \text{m} = 6.2 \, \text{mm}
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