The minimum incident angle would ray of light undergo total internal reflection.

Question

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Answer 1

Question

Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

(a)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

College Physics:, Chapter 22, Problem 40P , additional homework tip 1

Formula to calculate the minimum incident angle is,

$θi=sin−1(nasin(90)ng)$

Here,

$θi$ is the minimum incident angle

$na and ng$ are the refractive index of air and glass

Substitute $1$ for $na$ , $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)sin(90)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

(b)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

College Physics:, Chapter 22, Problem 40P , additional homework tip 2

Expressing the formula for the critical angle is,

$nwsinθc=nasin(90)sinθc=1nw$

Here,

$θc$ is the critical angle

$nw$ is the refractive index of water

$na$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$θi=sin−1(nwsinθcng)=sin−1(nw(1nw)ng)=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Substitute, $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

(c)

Expert Solution

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

(d)

Expert Solution

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.

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Answer 2

Question

Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

(a)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

College Physics:, Chapter 22, Problem 40P , additional homework tip 1

Formula to calculate the minimum incident angle is,

$θi=sin−1(nasin(90)ng)$

Here,

$θi$ is the minimum incident angle

$na and ng$ are the refractive index of air and glass

Substitute $1$ for $na$ , $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)sin(90)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

(b)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

College Physics:, Chapter 22, Problem 40P , additional homework tip 2

Expressing the formula for the critical angle is,

$nwsinθc=nasin(90)sinθc=1nw$

Here,

$θc$ is the critical angle

$nw$ is the refractive index of water

$na$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$θi=sin−1(nwsinθcng)=sin−1(nw(1nw)ng)=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Substitute, $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

(c)

Expert Solution

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

(d)

Expert Solution

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.

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Answer 3

Question

Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

(a)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

College Physics:, Chapter 22, Problem 40P , additional homework tip 1

Formula to calculate the minimum incident angle is,

$θi=sin−1(nasin(90)ng)$

Here,

$θi$ is the minimum incident angle

$na and ng$ are the refractive index of air and glass

Substitute $1$ for $na$ , $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)sin(90)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

(b)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

College Physics:, Chapter 22, Problem 40P , additional homework tip 2

Expressing the formula for the critical angle is,

$nwsinθc=nasin(90)sinθc=1nw$

Here,

$θc$ is the critical angle

$nw$ is the refractive index of water

$na$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$θi=sin−1(nwsinθcng)=sin−1(nw(1nw)ng)=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Substitute, $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

(c)

Expert Solution

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

(d)

Expert Solution

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.

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Answer 4

Question

Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

(a)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

College Physics:, Chapter 22, Problem 40P , additional homework tip 1

Formula to calculate the minimum incident angle is,

$θi=sin−1(nasin(90)ng)$

Here,

$θi$ is the minimum incident angle

$na and ng$ are the refractive index of air and glass

Substitute $1$ for $na$ , $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)sin(90)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

(b)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

College Physics:, Chapter 22, Problem 40P , additional homework tip 2

Expressing the formula for the critical angle is,

$nwsinθc=nasin(90)sinθc=1nw$

Here,

$θc$ is the critical angle

$nw$ is the refractive index of water

$na$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$θi=sin−1(nwsinθcng)=sin−1(nw(1nw)ng)=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Substitute, $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

(c)

Expert Solution

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

(d)

Expert Solution

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

(a)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

College Physics:, Chapter 22, Problem 40P , additional homework tip 1

Formula to calculate the minimum incident angle is,

$θi=sin−1(nasin(90)ng)$

Here,

$θi$ is the minimum incident angle

$na and ng$ are the refractive index of air and glass

Substitute $1$ for $na$ , $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)sin(90)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

(b)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

College Physics:, Chapter 22, Problem 40P , additional homework tip 2

Expressing the formula for the critical angle is,

$nwsinθc=nasin(90)sinθc=1nw$

Here,

$θc$ is the critical angle

$nw$ is the refractive index of water

$na$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$θi=sin−1(nwsinθcng)=sin−1(nw(1nw)ng)=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Substitute, $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

(c)

Expert Solution

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

(d)

Expert Solution

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.

Answer 6

Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

(a)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

College Physics:, Chapter 22, Problem 40P , additional homework tip 1

Formula to calculate the minimum incident angle is,

$θi=sin−1(nasin(90)ng)$

Here,

$θi$ is the minimum incident angle

$na and ng$ are the refractive index of air and glass

Substitute $1$ for $na$ , $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)sin(90)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

Answer 7

Chapter 22, Problem 40P

(a)

To determine

The minimum incident angle would ray of light undergo total internal reflection.

Answer 8

(a)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection is

$34.2°$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The ray diagram for the minimum incident angle is given below:

College Physics:, Chapter 22, Problem 40P , additional homework tip 1

Formula to calculate the minimum incident angle is,

$θi=sin−1(nasin(90)ng)$

Here,

$θi$ is the minimum incident angle

$na and ng$ are the refractive index of air and glass

Substitute $1$ for $na$ , $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)sin(90)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

Answer 13

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

(b)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is

$34.2°$ .

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

College Physics:, Chapter 22, Problem 40P , additional homework tip 2

Expressing the formula for the critical angle is,

$nwsinθc=nasin(90)sinθc=1nw$

Here,

$θc$ is the critical angle

$nw$ is the refractive index of water

$na$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$θi=sin−1(nwsinθcng)=sin−1(nw(1nw)ng)=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Substitute, $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

Answer 14

(b)

To determine

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

Answer 15

(b)

Expert Solution

Answer to Problem 40P

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass is

$34.2°$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The ray diagram for the minimum incident angle for the glass water interface is,

College Physics:, Chapter 22, Problem 40P , additional homework tip 2

Expressing the formula for the critical angle is,

$nwsinθc=nasin(90)sinθc=1nw$

Here,

$θc$ is the critical angle

$nw$ is the refractive index of water

$na$ is the refractive index of air

Formula to calculate the minimum incident angle is,

$θi=sin−1(nwsinθcng)=sin−1(nw(1nw)ng)=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Substitute, $1.78$ for $ng$ to find $θi$ ,

$θi=sin−1((1)(1.78))=34.2°$

Conclusion:

Therefore, The minimum incident angle would ray of light undergo total internal reflection is $34.2°$ .

Answer 20

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

(c)

Expert Solution

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Answer 21

(c)

To determine

The effect of the thickness of water layer or glass on the minimum incident angle.

Answer 22

(c)

Expert Solution

Answer to Problem 40P

The thickness of water layer or glass does not on the minimum incident angle.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Thus, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the thickness of the water layer of glass layer.

Answer 27

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

(d)

Expert Solution

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.

Answer 28

(d)

To determine

The effect of the refractive index of the intervening layer on the minimum incident angle.

Answer 29

(d)

Expert Solution

Answer to Problem 40P

The refractive index of the intervening layer does not on the minimum incident angle.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Expression for the minimum incident angle is,

$θi=sin−1(1ng)$

Here,

$θi$ is the minimum incident angle

$ng$ is the refractive index of glass

Conclusion:

Therefore, the minimum incident angle would ray of light undergo total internal reflection does not depend on the refractive index of the intervening layer.

Answer 34

Ch. 22.2 - Which part of Figure 22.3, (a) or (b), better...Ch. 22.2 - Prob. 22.2QQ Ch. 22.3 - A material has an index of refraction that...Ch. 22.3 - As light travels from a vacuum (n = 1) to a medium...Ch. 22 - Prob. 1CQ Ch. 22 - A ray of light passes from one material into a...Ch. 22 - Prob. 3CQ Ch. 22 - Prob. 4CQ Ch. 22 - Determine whether each of the following statements...Ch. 22 - A type of mirage called a pingo is often observed...

The minimum incident angle would ray of light undergo total internal reflection.

Concept explainers

The minimum incident angle would ray of light undergo total internal reflection.

Answer to Problem 40P

Explanation of Solution

The minimum incident angle would ray of light undergo total internal reflection if water is placed over the glass.

Answer to Problem 40P

Explanation of Solution

The effect of the thickness of water layer or glass on the minimum incident angle.

Answer to Problem 40P

Explanation of Solution

The effect of the refractive index of the intervening layer on the minimum incident angle.

Answer to Problem 40P

Explanation of Solution

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