The angle made by the incident light with the surface of index n 2 is tan θ = n 2 sin β n 1 − n 2 cos β .

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Answer 1

Question

Chapter 37, Problem 46AP

(a)

To determine

To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

(b)

To determine

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

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Answer 2

Question

Chapter 37, Problem 46AP

(a)

To determine

To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

(b)

To determine

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

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Answer 3

Question

Chapter 37, Problem 46AP

(a)

To determine

To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

(b)

To determine

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

Expert Solution & Answer

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Answer 4

Question

Chapter 37, Problem 46AP

(a)

To determine

To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

(b)

To determine

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

Expert Solution & Answer

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Answer 5

Chapter 37, Problem 46AP

(a)

To determine

To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

(b)

To determine

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

Answer 6

Chapter 37, Problem 46AP

(a)

To determine

To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

Answer 7

Chapter 37, Problem 46AP

(a)

To determine

To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

Answer 8

(b)

To determine

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

Answer 9

(b)

To determine

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

Answer 10

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Answer 11

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Answer 12

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Answer 13

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The angle made by the incident light with the surface of index n 2 is tan θ = n 2 sin β n 1 − n 2 cos β .

Solution Summary: The diagram for the given condition is shown below.

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To show: The angle made by the incident light with the surface of index n2 is tanθ=n2sinβn1−n2cosβ .

To show: The equation, tanθ=n2sinβn1−n2cosβ reduces to Brewster’s law when β=90° .

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Chapter 37 Solutions