12. Light of wavelength 532 nm and intensity Im illuminates a single slit 3.73 µm wide.Find the relative intensity (that is, the value of I/Im) of the diffraction pattern at an angle of5.70⁰ from the central maximum.

Answered: Light of wavelength 532 nm and…

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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12. Light of wavelength 532 nm and intensity Im illuminates a single slit 3.73 µm wide.
Find the relative intensity (that is, the value of I/Im) of the diffraction pattern at an angle of
5.70⁰ from the central maximum.

Expert Solution

Step 1

The given problem involves the concept of single-slit diffraction of light. When a beam of light passes through a narrow slit, it spreads out into a diffraction pattern consisting of a central bright maximum and alternating dark and bright fringes on either side. The angle at which the bright fringes are observed is related to the width of the slit and the wavelength of light.

The problem provides the following information:

Wavelength of light, $λ = 532 n m = 5.32 \times 10^{- 7} m$
Width of the slit, $a = 3.73 μ m = 3.73 \times 10^{- 6} m$
Angle from central maximum, θ = 5.70°

To find the relative intensity of the diffraction pattern at the given angle, we can use the following formula:

$I / I_{m} = (s i n {(β) / β)}^{2}$

where I/I_m is the relative intensity, and $β = \frac{πasin (θ)}{λ}$ .

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