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High School Math Algebra Holt Mcdougal Larson Algebra 2: Student Edition 2012

a. To state: The simplified form of the equation to derive Snell’s law: n 1 sin θ 1 = n 2 sin θ 2 The change of direction is defined by Snell’s law where n 1 and n 2 are the indices of refraction for the two mediums. Snell’s law can be derived from the equation: n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 The resultant answer is n 1 sin θ 1 = n 2 sin θ 2 . Given information: The given equation is: n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 Explanation: Consider the given equation: n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 Use Pythagorean identity cot 2 θ + 1 = csc 2 θ , n 1 csc 2 θ 1 = n 2 csc 2 θ 2 n 1 csc θ 1 = n 2 csc θ 2 Substitute 1 csc θ = sin θ , n 1 sin θ 1 = n 2 sin θ 2 Therefore, the simplified form of the equation n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 is n 1 sin θ 1 = n 2 sin θ 2 . b. To state: The value of n 1 if θ 1 = 55 ∘ , θ 2 = 35 ∘ , and n 2 = 2 . The resultant answer is m ∠ A = 37 ∘ . Given information: The equation from part (a) is n 1 sin θ 1 = n 2 sin θ 2 . Explanation: Consider the equation n 1 sin θ 1 = n 2 sin θ 2 , n 1 sin θ 1 = n 2 sin θ 2 Now substitute the values θ 1 = 55 ∘ , θ 2 = 35 ∘ , and n 2 = 2 . n 1 sin 55 ∘ = 2 sin 35 ∘ n 1 ⋅ 0.819 = 2 ⋅ 0.574 n 1 = 1.148 0.819 n 1 ≈ 1.402 Therefore, the value of n 1 is about 1.402 . c. To state: What must be true for n 1 and n 2 if θ 1 = θ 2 . If θ 1 = θ 2 then n 1 and n 2 must be equal. Given information: It is given that θ 1 = θ 2 . Explanation: Consider the equation n 1 sin θ 1 = n 2 sin θ 2 , It is given that θ 1 = θ 2 , then it can be written as: n 1 sin θ 1 = n 2 sin θ 1 n 1 = n 2 Here n 1 and n 2 represents indices of refraction. It means the indices of refraction for the two mediums are equal.

a. To state: The simplified form of the equation to derive Snell’s law: n 1 sin θ 1 = n 2 sin θ 2 The change of direction is defined by Snell’s law where n 1 and n 2 are the indices of refraction for the two mediums. Snell’s law can be derived from the equation: n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 The resultant answer is n 1 sin θ 1 = n 2 sin θ 2 . Given information: The given equation is: n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 Explanation: Consider the given equation: n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 Use Pythagorean identity cot 2 θ + 1 = csc 2 θ , n 1 csc 2 θ 1 = n 2 csc 2 θ 2 n 1 csc θ 1 = n 2 csc θ 2 Substitute 1 csc θ = sin θ , n 1 sin θ 1 = n 2 sin θ 2 Therefore, the simplified form of the equation n 1 cot 2 θ 1 + 1 = n 2 cot 2 θ 2 + 1 is n 1 sin θ 1 = n 2 sin θ 2 . b. To state: The value of n 1 if θ 1 = 55 ∘ , θ 2 = 35 ∘ , and n 2 = 2 . The resultant answer is m ∠ A = 37 ∘ . Given information: The equation from part (a) is n 1 sin θ 1 = n 2 sin θ 2 . Explanation: Consider the equation n 1 sin θ 1 = n 2 sin θ 2 , n 1 sin θ 1 = n 2 sin θ 2 Now substitute the values θ 1 = 55 ∘ , θ 2 = 35 ∘ , and n 2 = 2 . n 1 sin 55 ∘ = 2 sin 35 ∘ n 1 ⋅ 0.819 = 2 ⋅ 0.574 n 1 = 1.148 0.819 n 1 ≈ 1.402 Therefore, the value of n 1 is about 1.402 . c. To state: What must be true for n 1 and n 2 if θ 1 = θ 2 . If θ 1 = θ 2 then n 1 and n 2 must be equal. Given information: It is given that θ 1 = θ 2 . Explanation: Consider the equation n 1 sin θ 1 = n 2 sin θ 2 , It is given that θ 1 = θ 2 , then it can be written as: n 1 sin θ 1 = n 2 sin θ 1 n 1 = n 2 Here n 1 and n 2 represents indices of refraction. It means the indices of refraction for the two mediums are equal.

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Holt Mcdougal Larson Algebra 2: Student Edition 2012

1st Edition

ISBN: 9780547647159

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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Expert Solution & Answer

Chapter 10.3, Problem 44PS

Solution

a.

To state: The simplified form of the equation to derive Snell’s law: n1sinθ1=n2sinθ2

The change of direction is defined by Snell’s law where n1 and n2 are the indices of refraction for the two mediums. Snell’s law can be derived from the equation:

n1cot2θ1+1=n2cot2θ2+1

The resultant answer is n1sinθ1=n2sinθ2 .

Given information:

The given equation is:

n1cot2θ1+1=n2cot2θ2+1

Explanation:

Consider the given equation:

n1cot2θ1+1=n2cot2θ2+1

Use Pythagorean identity cot2θ+1=csc2θ ,

n1csc2θ1=n2csc2θ2n1cscθ1=n2cscθ2

Substitute 1cscθ=sinθ ,

n1sinθ1=n2sinθ2

Therefore, the simplified form of the equation n1cot2θ1+1=n2cot2θ2+1 is n1sinθ1=n2sinθ2 .

b.

To state: The value of n1 if θ1=55∘, θ2=35∘, and n2=2 .

The resultant answer is m∠A=37∘ .

Given information:

The equation from part (a) is n1sinθ1=n2sinθ2 .

Explanation:

Consider the equation n1sinθ1=n2sinθ2 ,

n1sinθ1=n2sinθ2

Now substitute the values θ1=55∘, θ2=35∘, and n2=2 .

n1sin55∘=2sin35∘n1⋅0.819=2⋅0.574n1=1.1480.819n1≈1.402

Therefore, the value of n1 is about 1.402 .

c.

To state: What must be true for n1 and n2 if θ1=θ2 .

If θ1=θ2 then n1 and n2 must be equal.

Given information:

It is given that θ1=θ2 .

Explanation:

Consider the equation n1sinθ1=n2sinθ2 ,

It is given that θ1=θ2 , then it can be written as:

n1sinθ1=n2sinθ1n1=n2

Here n1 and n2 represents indices of refraction. It means the indices of refraction for the two mediums are equal.

Chapter 10.3, Problem 43PS

Chapter 10.3, Problem 45PS

Chapter 10 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 10.1 - Prob. 1GP Ch. 10.1 - Prob. 2GP Ch. 10.1 - Prob. 3GP Ch. 10.1 - Prob. 4GP Ch. 10.1 - Prob. 5GP Ch. 10.1 - Prob. 6GP Ch. 10.1 - Prob. 7GP Ch. 10.1 - Prob. 8GP Ch. 10.1 - Prob. 9GP Ch. 10.1 - Prob. 10GP

Ch. 10.1 - Prob. 11GP Ch. 10.1 - Prob. 12GP Ch. 10.1 - Prob. 13GP Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29PS Ch. 10.1 - Prob. 30PS Ch. 10.1 - Prob. 31PS Ch. 10.1 - Prob. 32PS Ch. 10.1 - Prob. 33PS Ch. 10.2 - Prob. 1GP Ch. 10.2 - Prob. 2GP Ch. 10.2 - Prob. 3GP Ch. 10.2 - Prob. 4GP Ch. 10.2 - Prob. 5GP Ch. 10.2 - Prob. 6GP Ch. 10.2 - Prob. 7GP Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - 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