The angle of refraction θ s in the sapphire crystal.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction $θs$ in the sapphire crystal.

(a)

Expert Solution

Answer to Problem 96PQ

The angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$nasinθa=nssinθs$ (I)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $θs$ .

$nasinθa=nssinθssinθs=sinθa(nans)θs=sin−1(sinθa(nans))$ (II)

Conclusion:

Substitute 1.760 for $ns$ , $27.0°$ for $θa$ , $1.00$ for $na$ in equation (II).

Here, $ns=1.495$ and $θa=27.0°$ are taken from given data and $na=1.00$ for the air medium.

$θs=sin−1(sinθa(nans))=sin−1(sin27.0°(1.001.760))=sin−1(0.45390(0.56818))=15.0°$

Therefore, the angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

(b)

Expert Solution

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Explanation of Solution

Consider the diagram.

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term, Chapter 38, Problem 96PQ

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is $θa=27.0°$ and the angle of refraction of the light ray inside the crystal is $θs$ .

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface $θs$ is equal to the angle of incidence at the sapphire interface $θa$ .

$θs=θa$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

(c)

Expert Solution

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$na,1sinθa,1=ns,1sinθs,1$ (III)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $sinθs,1$ .

$na,1sinθa,1=ns,1sinθs,1sinθs,1=sinθa,1(na,1ns,1)θs,1=sin−1(sinθa,1(na,1ns,1))$ (IV)

Conclusion:

Substitute $1.00$ for $ns,1$ , $14.94°$ for $θa,1$ ,1.760 for $na,1$ in equation (IV).

$θs,1=sin−1(sinθa,1(na,1ns,1))=sin−1(sin14.94°(1.7601.00))=sin−1((0.2578)(1.760))=27.0°$

Therefore, the angle of refraction of the light ray at the crystal air interface is $27.0°$ .

Want to see more full solutions like this?

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

Students have asked these similar questions

••63 SSM www In the circuit of Fig. 27-65, 8 = 1.2 kV, C = 6.5 µF, R₁ S R₂ R3 800 C H R₁ = R₂ = R3 = 0.73 MQ. With C completely uncharged, switch S is suddenly closed (at t = 0). At t = 0, what are (a) current i̟ in resistor 1, (b) current 2 in resistor 2, and (c) current i3 in resistor 3? At t = ∞o (that is, after many time constants), what are (d) i₁, (e) i₂, and (f) iz? What is the potential difference V2 across resistor 2 at (g) t = 0 and (h) t = ∞o? (i) Sketch V2 versus t between these two extreme times. Figure 27-65 Problem 63.

Thor flies by spinning his hammer really fast from a leather strap at the end of the handle, letting go, then grabbing it and having it pull him. If Thor wants to reach escape velocity (velocity needed to leave Earth’s atmosphere), he will need the linear velocity of the center of mass of the hammer to be 11,200 m/s. Thor's escape velocity is 33532.9 rad/s, the angular velocity is 8055.5 rad/s^2. While the hammer is spinning at its maximum speed what impossibly large tension does the leather strap, which the hammer is spinning by, exert when the hammer is at its lowest point? the hammer has a total mass of 20.0kg.

If the room’s radius is 16.2 m, at what minimum linear speed does Quicksilver need to run to stay on the walls without sliding down? Assume the coefficient of friction between Quicksilver and the wall is 0.236.

Answer 2

Question

Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction $θs$ in the sapphire crystal.

(a)

Expert Solution

Answer to Problem 96PQ

The angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$nasinθa=nssinθs$ (I)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $θs$ .

$nasinθa=nssinθssinθs=sinθa(nans)θs=sin−1(sinθa(nans))$ (II)

Conclusion:

Substitute 1.760 for $ns$ , $27.0°$ for $θa$ , $1.00$ for $na$ in equation (II).

Here, $ns=1.495$ and $θa=27.0°$ are taken from given data and $na=1.00$ for the air medium.

$θs=sin−1(sinθa(nans))=sin−1(sin27.0°(1.001.760))=sin−1(0.45390(0.56818))=15.0°$

Therefore, the angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

(b)

Expert Solution

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Explanation of Solution

Consider the diagram.

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term, Chapter 38, Problem 96PQ

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is $θa=27.0°$ and the angle of refraction of the light ray inside the crystal is $θs$ .

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface $θs$ is equal to the angle of incidence at the sapphire interface $θa$ .

$θs=θa$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

(c)

Expert Solution

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$na,1sinθa,1=ns,1sinθs,1$ (III)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $sinθs,1$ .

$na,1sinθa,1=ns,1sinθs,1sinθs,1=sinθa,1(na,1ns,1)θs,1=sin−1(sinθa,1(na,1ns,1))$ (IV)

Conclusion:

Substitute $1.00$ for $ns,1$ , $14.94°$ for $θa,1$ ,1.760 for $na,1$ in equation (IV).

$θs,1=sin−1(sinθa,1(na,1ns,1))=sin−1(sin14.94°(1.7601.00))=sin−1((0.2578)(1.760))=27.0°$

Therefore, the angle of refraction of the light ray at the crystal air interface is $27.0°$ .

Want to see more full solutions like this?

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

Answer 3

Question

Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction $θs$ in the sapphire crystal.

(a)

Expert Solution

Answer to Problem 96PQ

The angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$nasinθa=nssinθs$ (I)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $θs$ .

$nasinθa=nssinθssinθs=sinθa(nans)θs=sin−1(sinθa(nans))$ (II)

Conclusion:

Substitute 1.760 for $ns$ , $27.0°$ for $θa$ , $1.00$ for $na$ in equation (II).

Here, $ns=1.495$ and $θa=27.0°$ are taken from given data and $na=1.00$ for the air medium.

$θs=sin−1(sinθa(nans))=sin−1(sin27.0°(1.001.760))=sin−1(0.45390(0.56818))=15.0°$

Therefore, the angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

(b)

Expert Solution

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Explanation of Solution

Consider the diagram.

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term, Chapter 38, Problem 96PQ

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is $θa=27.0°$ and the angle of refraction of the light ray inside the crystal is $θs$ .

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface $θs$ is equal to the angle of incidence at the sapphire interface $θa$ .

$θs=θa$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

(c)

Expert Solution

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$na,1sinθa,1=ns,1sinθs,1$ (III)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $sinθs,1$ .

$na,1sinθa,1=ns,1sinθs,1sinθs,1=sinθa,1(na,1ns,1)θs,1=sin−1(sinθa,1(na,1ns,1))$ (IV)

Conclusion:

Substitute $1.00$ for $ns,1$ , $14.94°$ for $θa,1$ ,1.760 for $na,1$ in equation (IV).

$θs,1=sin−1(sinθa,1(na,1ns,1))=sin−1(sin14.94°(1.7601.00))=sin−1((0.2578)(1.760))=27.0°$

Therefore, the angle of refraction of the light ray at the crystal air interface is $27.0°$ .

Want to see more full solutions like this?

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

Answer 4

Question

Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction $θs$ in the sapphire crystal.

(a)

Expert Solution

Answer to Problem 96PQ

The angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$nasinθa=nssinθs$ (I)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $θs$ .

$nasinθa=nssinθssinθs=sinθa(nans)θs=sin−1(sinθa(nans))$ (II)

Conclusion:

Substitute 1.760 for $ns$ , $27.0°$ for $θa$ , $1.00$ for $na$ in equation (II).

Here, $ns=1.495$ and $θa=27.0°$ are taken from given data and $na=1.00$ for the air medium.

$θs=sin−1(sinθa(nans))=sin−1(sin27.0°(1.001.760))=sin−1(0.45390(0.56818))=15.0°$

Therefore, the angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

(b)

Expert Solution

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Explanation of Solution

Consider the diagram.

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term, Chapter 38, Problem 96PQ

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is $θa=27.0°$ and the angle of refraction of the light ray inside the crystal is $θs$ .

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface $θs$ is equal to the angle of incidence at the sapphire interface $θa$ .

$θs=θa$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

(c)

Expert Solution

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$na,1sinθa,1=ns,1sinθs,1$ (III)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $sinθs,1$ .

$na,1sinθa,1=ns,1sinθs,1sinθs,1=sinθa,1(na,1ns,1)θs,1=sin−1(sinθa,1(na,1ns,1))$ (IV)

Conclusion:

Substitute $1.00$ for $ns,1$ , $14.94°$ for $θa,1$ ,1.760 for $na,1$ in equation (IV).

$θs,1=sin−1(sinθa,1(na,1ns,1))=sin−1(sin14.94°(1.7601.00))=sin−1((0.2578)(1.760))=27.0°$

Therefore, the angle of refraction of the light ray at the crystal air interface is $27.0°$ .

Want to see more full solutions like this?

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

Answer 5

Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction $θs$ in the sapphire crystal.

(a)

Expert Solution

Answer to Problem 96PQ

The angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$nasinθa=nssinθs$ (I)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $θs$ .

$nasinθa=nssinθssinθs=sinθa(nans)θs=sin−1(sinθa(nans))$ (II)

Conclusion:

Substitute 1.760 for $ns$ , $27.0°$ for $θa$ , $1.00$ for $na$ in equation (II).

Here, $ns=1.495$ and $θa=27.0°$ are taken from given data and $na=1.00$ for the air medium.

$θs=sin−1(sinθa(nans))=sin−1(sin27.0°(1.001.760))=sin−1(0.45390(0.56818))=15.0°$

Therefore, the angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

(b)

Expert Solution

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Explanation of Solution

Consider the diagram.

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term, Chapter 38, Problem 96PQ

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is $θa=27.0°$ and the angle of refraction of the light ray inside the crystal is $θs$ .

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface $θs$ is equal to the angle of incidence at the sapphire interface $θa$ .

$θs=θa$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

(c)

Expert Solution

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$na,1sinθa,1=ns,1sinθs,1$ (III)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $sinθs,1$ .

$na,1sinθa,1=ns,1sinθs,1sinθs,1=sinθa,1(na,1ns,1)θs,1=sin−1(sinθa,1(na,1ns,1))$ (IV)

Conclusion:

Substitute $1.00$ for $ns,1$ , $14.94°$ for $θa,1$ ,1.760 for $na,1$ in equation (IV).

$θs,1=sin−1(sinθa,1(na,1ns,1))=sin−1(sin14.94°(1.7601.00))=sin−1((0.2578)(1.760))=27.0°$

Therefore, the angle of refraction of the light ray at the crystal air interface is $27.0°$ .

Answer 6

Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction $θs$ in the sapphire crystal.

(a)

Expert Solution

Answer to Problem 96PQ

The angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$nasinθa=nssinθs$ (I)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $θs$ .

$nasinθa=nssinθssinθs=sinθa(nans)θs=sin−1(sinθa(nans))$ (II)

Conclusion:

Substitute 1.760 for $ns$ , $27.0°$ for $θa$ , $1.00$ for $na$ in equation (II).

Here, $ns=1.495$ and $θa=27.0°$ are taken from given data and $na=1.00$ for the air medium.

$θs=sin−1(sinθa(nans))=sin−1(sin27.0°(1.001.760))=sin−1(0.45390(0.56818))=15.0°$

Therefore, the angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Answer 7

Chapter 38, Problem 96PQ

(a)

To determine

The angle of refraction $θs$ in the sapphire crystal.

Answer 8

(a)

Expert Solution

Answer to Problem 96PQ

The angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$nasinθa=nssinθs$ (I)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $θs$ .

$nasinθa=nssinθssinθs=sinθa(nans)θs=sin−1(sinθa(nans))$ (II)

Conclusion:

Substitute 1.760 for $ns$ , $27.0°$ for $θa$ , $1.00$ for $na$ in equation (II).

Here, $ns=1.495$ and $θa=27.0°$ are taken from given data and $na=1.00$ for the air medium.

$θs=sin−1(sinθa(nans))=sin−1(sin27.0°(1.001.760))=sin−1(0.45390(0.56818))=15.0°$

Therefore, the angle of refraction $θs$ , in the sapphire crystal is $15.0°$ .

Answer 13

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

(b)

Expert Solution

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Explanation of Solution

Consider the diagram.

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term, Chapter 38, Problem 96PQ

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is $θa=27.0°$ and the angle of refraction of the light ray inside the crystal is $θs$ .

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface $θs$ is equal to the angle of incidence at the sapphire interface $θa$ .

$θs=θa$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Answer 14

(b)

To determine

The angle of incidence of the light ray at the sapphire air interface.

Answer 15

(b)

Expert Solution

Answer to Problem 96PQ

The angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Consider the diagram.

Webassign Printed Access Card For Katz's Physics For Scientists And Engineers: Foundations And Connections, 1st Edition, Single-term, Chapter 38, Problem 96PQ

Figure-(1)

The above figure defines the light ray incident on the rectangular sapphire crystal at an angle of incidence is $θa=27.0°$ and the angle of refraction of the light ray inside the crystal is $θs$ .

Inside the sapphire crystal the ray travels in a straight line because the medium is same. So the angle of refraction at the air sapphire interface $θs$ is equal to the angle of incidence at the sapphire interface $θa$ .

$θs=θa$

Therefore, the angle of incidence of the light ray at the sapphire air interface is $15.0°$ .

Answer 20

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

(c)

Expert Solution

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$ .

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$na,1sinθa,1=ns,1sinθs,1$ (III)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.

Solve the above equation to find $sinθs,1$ .

$na,1sinθa,1=ns,1sinθs,1sinθs,1=sinθa,1(na,1ns,1)θs,1=sin−1(sinθa,1(na,1ns,1))$ (IV)

Conclusion:

Substitute $1.00$ for $ns,1$ , $14.94°$ for $θa,1$ ,1.760 for $na,1$ in equation (IV).

$θs,1=sin−1(sinθa,1(na,1ns,1))=sin−1(sin14.94°(1.7601.00))=sin−1((0.2578)(1.760))=27.0°$

Therefore, the angle of refraction of the light ray at the crystal air interface is $27.0°$ .

Answer 21

(c)

To determine

The angle of refraction of the light ray at crystal air interface.

Answer 22

(c)

Expert Solution

Answer to Problem 96PQ

The angle of refraction of the light ray at crystal air interface is $27.0°$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for Snell’s law that relates the angle of incidence and angle of refraction of a light ray when it travels from one medium to the other medium.

$na,1sinθa,1=ns,1sinθs,1$ (III)

Here, $na$ is the refractive index of the air medium, $ns$ is the refractive index of the sapphire crystal medium, $θa$ is the angle of incidence of air and $θs$ is the angle of refraction of sapphire crystal.