### Diffraction Pattern Analysis**Problem Statement:**Richard observes that a single slit that is 2,022 nm wide forms a diffraction pattern when illuminated by monochromatic light of 673 nm wavelength. At an angle of 10° from the central maximum, what is the ratio of the intensity to the intensity of the central maximum? Give your answer with two decimal places, please.**Solution:**To determine the ratio of the intensity at an angle (θ) to the intensity at the central maximum for a single slit diffraction pattern, we use the diffraction intensity formula:\[ I(\theta) = I_0 \left(\frac{\sin(\beta/2)}{\beta/2}\right)^2 \]where:- $ \beta = \frac{2 \pi a \sin(\theta)}{\lambda} $- $ a $ is the width of the slit- $ \lambda $ is the wavelength of the light- $ \theta $ is the angle from the central maximum- $ I_0 $ is the central maximum intensity.Given:- $ a = 2022 $ nm- $ \lambda = 673 $ nm- $ \theta = 10° $We first calculate $ \beta $:\[ \beta = \frac{2 \pi \times 2022 \times \sin(10°)}{673} \]\[ \sin(10°) \approx 0.1736 \]\[ \beta = \frac{2 \pi \times 2022 \times 0.1736}{673} \]\[ \beta \approx \frac{2201.584 \times 0.1736}{673} \]\[ \beta \approx 0.5699 \]Now plug $ \beta $ into the intensity formula:\[ I(\theta) = I_0 \left(\frac{\sin(0.5699/2)}{0.5699/2}\right)^2 \]\[ I(\theta) = I_0 \left(\frac{\sin(0.28495)}{0.28495}\right)^2 \]\[ \sin(0.28495) \approx 0.2810 \]\[ I(\theta) = I_0 \left(\frac{0.2810}{0.28495}\right)^2 \]\[ \

Given: The width of the slit, a = 2022 nm wavelength of light, λ = 673 nm angle from central maxima,…

Richard observes that a single slit that is2,022 nm wide forms a diffraction pattern when illuminated by monochromatic light of673 -nm wavelength. At an angle of 10° from the central maximum, what is the ratio of the intensity to the intensity of the central maximum? Give your answer with two decimal places please.

Richard observes that a single slit that is2,022 nm wide forms a diffraction pattern when illuminated by monochromatic light of673 -nm wavelength. At an angle of 10° from the central maximum, what is the ratio of the intensity to the intensity of the central maximum? Give your answer with two 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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Concept explainers

Diffraction of light

In physics, the term diffraction is defined as the bending of light around the corners of the obstacle. As a light ray passes by an aperture, it bends around the edges or through slits, it spreads out through a narrow opening. The diffracted light will produce fringes of bright and dark or colored bands which are labeled as a diffraction pattern.

Question

### Diffraction Pattern Analysis

**Problem Statement:**
Richard observes that a single slit that is 2,022 nm wide forms a diffraction pattern when illuminated by monochromatic light of 673 nm wavelength. At an angle of 10° from the central maximum, what is the ratio of the intensity to the intensity of the central maximum? Give your answer with two decimal places, please.

**Solution:**

To determine the ratio of the intensity at an angle (θ) to the intensity at the central maximum for a single slit diffraction pattern, we use the diffraction intensity formula:

\[ I(\theta) = I_0 \left(\frac{\sin(\beta/2)}{\beta/2}\right)^2 \]

where:
- $ \beta = \frac{2 \pi a \sin(\theta)}{\lambda} $
- $ a $ is the width of the slit
- $ \lambda $ is the wavelength of the light
- $ \theta $ is the angle from the central maximum
- $ I_0 $ is the central maximum intensity.

Given:
- $ a = 2022 $ nm
- $ \lambda = 673 $ nm
- $ \theta = 10° $

We first calculate $ \beta $:

\[ \beta = \frac{2 \pi \times 2022 \times \sin(10°)}{673} \]

\[ \sin(10°) \approx 0.1736 \]

\[ \beta = \frac{2 \pi \times 2022 \times 0.1736}{673} \]

\[ \beta \approx \frac{2201.584 \times 0.1736}{673} \]

\[ \beta \approx 0.5699 \]

Now plug $ \beta $ into the intensity formula:

\[ I(\theta) = I_0 \left(\frac{\sin(0.5699/2)}{0.5699/2}\right)^2 \]

\[ I(\theta) = I_0 \left(\frac{\sin(0.28495)}{0.28495}\right)^2 \]

\[ \sin(0.28495) \approx 0.2810 \]

\[ I(\theta) = I_0 \left(\frac{0.2810}{0.28495}\right)^2 \]

\[ \

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