Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source a

Question

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply.

a) If the first diffraction minima are at ±90.0∘, so the central maximum completely fills the screen, what is the width of the slit?

b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at θ=45.0∘ to the intensity at θ=0?

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Thành Đạt 5 years 2021-08-19T19:18:07+00:00 1 Answers 30 views 0

Answers ( )

  1. MichaelMet
    0
    2021-08-19T19:20:01+00:00
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    Answer:

    Part a)

    Width of the slit is

    a = 580 nm

    Part b)

    Ratio of intensity is given as

    \frac{I}{I_o} = 0.81

    Explanation:

    Part a)

    As we know by the formula of diffraction we will have

    a sin\theta = \lambda

    so we have

    \theta = 90

    \lambda = 580 nm

    so we will have

    a sin90 = 580 nm

    a = 580 nm

    Part b)

    As we know that the intensity in diffraction pattern is given as

    I = I_o (\frac{sin\theta}{\theta})^2

    \frac{I}{I_o} = (\frac{sin\theta}{\theta})^2

    so for angle 45 degree

    \frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

    \frac{I}{I_o} = (\frac{sin45}{\pi/4})^2

    \frac{I}{I_o} = 0.81

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