Explanation 1) Concept: Here, we need to use the relation of the path length difference for dark fringes as the rays should be out of phase. For the protein solution, we need to consider the wavelength of light in the protein solution. 2) Formula: i) ∆ L = m + 1 2 λ ii) λ n = λ n 3) Given: i) λ = 500 n m ii) n = 1.38 4) Calculations: a) From the figure, the path length traveled by ray 1 is L 1 = 7 d And the path length traveled by ray 2 is L 2 = 2 d Then, the path length difference between ray 1 and 2 is L 1 - L 2 = 7 d - 2 d = 5 d To find value d at which ray 1 and ray 2 are out of phase, 5 d = λ 2

10th Edition

ISBN: 9781118230725

Author: David Halliday, Robert Resnick, Jearl Walker

Publisher: Wiley, John & Sons, Incorporated

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Chapter 35, Problem 83P

To determine

To find

a) For air medium, minimum value of d so that the rays are out of phase.

b) For protein solution, minimum value of d so that the rays are out of phase.

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Chapter 35, Problem 82P

Chapter 35, Problem 84P

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Chapter 35 Solutions

Fundamentals of Physics Extended

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To find a) For air medium, minimum value of d so that the rays are out of phase. b) For protein solution, minimum value of d so that the rays are out of phase.

Solution Summary: The author calculates the path length difference between rays 1 and 2, and the wavelength of light in the protein solution.

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To find a) For air medium, minimum value of d so that the rays are out of phase. b) For protein solution, minimum value of d so that the rays are out of phase.

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Chapter 35 Solutions

To find

a) For air medium, minimum value of d so that the rays are out of phase.

b) For protein solution, minimum value of d so that the rays are out of phase.