Light of wavelength 427 nm illuminates a round 0.2-mm diameter hole. A screen is placed 7.6 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? Give the answer with no decimal places please.
Light of wavelength 427 nm illuminates a round 0.2-mm diameter hole. A screen is placed 7.6 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? Give the answer with no decimal places please.
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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![**Determining the Diameter of the Central Bright Area in Light Diffraction**
**Problem Statement:**
Light of wavelength 427 nm illuminates a round 0.2-mm diameter hole. A screen is placed 7.6 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? Give the answer with no decimal places please.
**Explanation:**
When light of a specific wavelength passes through a small circular opening, it diffracts and forms a pattern on the screen. In this scenario, we have:
- **Wavelength (\(\lambda\))**: 427 nm (nanometers)
- **Diameter of the hole**: 0.2 mm (millimeters)
- **Distance from the hole to the screen (\(L\))**: 7.6 m (meters)
Using the formula for the diameter of the first diffraction minimum (central bright area) in the Fraunhofer diffraction pattern for a circular aperture:
\[ d = 2.44 \times \frac{\lambda L}{D} \]
Where:
- \(d\) is the diameter of the central bright area,
- \(\lambda\) is the wavelength of light,
- \(L\) is the distance from the hole to the screen,
- \(D\) is the diameter of the hole.
First, let's convert the wavelength into millimeters to be consistent:
\[ \lambda = 427 \, \text{nm} = 427 \times 10^{-6} \, \text{mm} \]
Plug these values into the formula to find \(d\):
\[ d = 2.44 \times \frac{427 \times 10^{-6} \times 7.6}{0.2} \]
Calculate the value of \(d\):
\[ d = 2.44 \times \frac{3245.2 \times 10^{-6}}{0.2} \]
\[ d = 2.44 \times 16226 \times 10^{-6} \]
\[ d = 2.44 \times 16.226 \]
\[ d \approx 39.6 \, \text{mm} \]
Rounding to the nearest whole number:
\[ \boxed{40 \, \text{mm}} \]
Therefore, the diameter of the central bright area on the screen is approximately **40 millimeters**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F98ea0ed6-be6f-4556-80a5-46c7ae10c3b9%2F707e7d76-1913-4934-82d4-8fcf6c9211b4%2F64941qc_processed.png&w=3840&q=75)
Transcribed Image Text:**Determining the Diameter of the Central Bright Area in Light Diffraction**
**Problem Statement:**
Light of wavelength 427 nm illuminates a round 0.2-mm diameter hole. A screen is placed 7.6 m behind the slit. What is the diameter of the central bright area on the screen in millimeters? Give the answer with no decimal places please.
**Explanation:**
When light of a specific wavelength passes through a small circular opening, it diffracts and forms a pattern on the screen. In this scenario, we have:
- **Wavelength (\(\lambda\))**: 427 nm (nanometers)
- **Diameter of the hole**: 0.2 mm (millimeters)
- **Distance from the hole to the screen (\(L\))**: 7.6 m (meters)
Using the formula for the diameter of the first diffraction minimum (central bright area) in the Fraunhofer diffraction pattern for a circular aperture:
\[ d = 2.44 \times \frac{\lambda L}{D} \]
Where:
- \(d\) is the diameter of the central bright area,
- \(\lambda\) is the wavelength of light,
- \(L\) is the distance from the hole to the screen,
- \(D\) is the diameter of the hole.
First, let's convert the wavelength into millimeters to be consistent:
\[ \lambda = 427 \, \text{nm} = 427 \times 10^{-6} \, \text{mm} \]
Plug these values into the formula to find \(d\):
\[ d = 2.44 \times \frac{427 \times 10^{-6} \times 7.6}{0.2} \]
Calculate the value of \(d\):
\[ d = 2.44 \times \frac{3245.2 \times 10^{-6}}{0.2} \]
\[ d = 2.44 \times 16226 \times 10^{-6} \]
\[ d = 2.44 \times 16.226 \]
\[ d \approx 39.6 \, \text{mm} \]
Rounding to the nearest whole number:
\[ \boxed{40 \, \text{mm}} \]
Therefore, the diameter of the central bright area on the screen is approximately **40 millimeters**.
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