Explanation 1) Concept: By using Bragg’s diffraction condition and from the given graph, we can find the shorter and the longer wavelength in the beam. 2) Formula: 2 d sin ⁡ θ = m λ 3) Given: i) Interplanar spacing, d = 0.94 n m ii) Horizontal scale, θ s = 2.00 ° iii) Order, m = 1 4) Calculations: a) By considering the graph from Fig. 36-46 and using the given horizontal scale, we can estimate that the first peak is at an angle 0.75 ° . So by Bragg’s diffraction condition for the first order, we can find the shorter wavelength of the beam. 2 d sin ⁡ θ 1 = λ 1 λ 1 = 2 × 0.94 × sin ⁡ 0.75 0 λ 1 = 2 × 0

10th Edition

ISBN: 9781118230718

Author: David Halliday

Publisher: Wiley, John & Sons, Incorporated

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Chapter 36, Problem 67P

To determine

To find:

a) The shorter wavelength in the beam

b) The longer wavelength in the beam

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Chapter 36, Problem 66P

Chapter 36, Problem 68P

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Chapter 36 Solutions

Fundamentals of Physics

Ch. 36 - You are conducting a single-slit diffraction...Ch. 36 - In a single-slit diffraction experiment, the top...Ch. 36 - For three experiments, Fig. 36-30 gives the...Ch. 36 - For three experiments, Fig. 36-31 gives versus...Ch. 36 - Figure 36-32 shows four choices for the...Ch. 36 - Prob. 6Q Ch. 36 - At night many people see rings called entoptic...Ch. 36 - a For a given diffraction grating, does the...Ch. 36 - Figure 36-33 shows a red line and a green line of...Ch. 36 - For the situation of Question 9 and Fig. 36-33, if...

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a) The shorter wavelength in the beam b) The longer wavelength in the beam

Solution Summary: The author explains Bragg's diftion condition and the given graph to find the shorter and longer wavelengths in the beam.

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To find: a) The shorter wavelength in the beam b) The longer wavelength in the beam

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Chapter 36 Solutions

To find:

a) The shorter wavelength in the beam

b) The longer wavelength in the beam