EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100454897
Author: Jewett
Publisher: YUZU
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Chapter 35, Problem 35.52AP

Consider a horizontal interface between air above and glass of index of refraction 1.55 below. (a) Draw a light ray incident from the air at angle of incidence 30.0°. Determine the angles of the reflected and refracted rays and show them on the diagram. (b) What If? Now suppose the light ray is incident from the glass at an angle of 30.0°. Determine the angles of the reflected and refracted rays and show all three rays on a new diagram. (c) For rays incident from the air onto the air–glass surface, determine and tabulate the angles of reflection and refraction for all the angles of incidence at 10.0° intervals from 0° to 90.0°. (d) Do the same for light rays coming up to the interface through the glass.

(a)

Expert Solution
Check Mark
To determine
The angle of the reflected and refracted rays when light ray incident from the air at angle of incidence 30° and show them on diagram.

Answer to Problem 35.52AP

The angle of reflection is 30° and the angle of refraction is 18.81° .

Explanation of Solution

Given info: The index of refraction of glass is 1.55 and the angle of incidence is 30° .

From, law of reflection, the angle of incidence is equal to the angle of reflection.

Thus, angle of reflection is 30° .

From Snell’s law,

n1sinθ1=n2sinθ2 (1)

Here,

θ1 is the angle of incidence.

θ2 is the angle of refraction.

n1 is the index of refraction of air.

n2 is the index of refraction of glass.

Substitute 1 for n1 , and 1.55 for n2 , 30° for θ1 , in equation(1) to calculate θ2 .

1×sin30°=1.55sinθ2sinθ2=sin30°1.55θ2=sin1[(12)1.55]=18.81°

The figure below shows the angle of incidence, the angle of reflection and the angle of refraction.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.52AP , additional homework tip  1

Figure (1)

Conclusion:

Therefore, the angle of reflection is 30° and the angle of refraction is 18.81° .

(b)

Expert Solution
Check Mark
To determine
The angle of the reflected and refracted rays when light ray incident from the glass at angle of incidence 30° and show them on diagram

Answer to Problem 35.52AP

The angle of reflection is 30° and the angle of refraction is 50.80° .

Explanation of Solution

Given info: The index of refraction of glass is 1.55 and angle of incidence is 30° .

From, law of reflection, angle of incidence is equal to angle of reflection.

Thus, angle of reflection is 30° .

From, Snell’s law,

n1sinθ1=n2sinθ2 (2)

Here,

n1 is the index of refraction of glass.

n2 is the index of refraction of air.

Substitute 1.55 for n1 , 1 for n2 , 30° for θ1 , in equation(2) to calculate θ2 .

1.55×sin30°=1×sinθ2sinθ2=1.55×12θ2=sin1(0.775)=50.80°

The figure below shows the angle of incidence, the angle of reflection and the angle of refraction.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 35, Problem 35.52AP , additional homework tip  2

Figure (2)

Conclusion:

Therefore, the angle of reflection is 30° and the angle of refraction is 50.80° .

(c)

Expert Solution
Check Mark
To determine
The angles of reflection and refraction for all angles of incidence at 10.0° from 0° to 90° in a tabular form for rays incident from air onto the air-glass surface.

Answer to Problem 35.52AP

The angle of reflection is same as the angle of incidence and the angle of refraction for rays incident from air onto the air-glass surface is increasing as angle of incidence is increasing. The table for different values of angles is,

Angle of incidence Angle of reflection Angle of refraction
0 0 0
10 10 6.43°
20 20 12.74°
30 30 18.81°
40 40 24.5°
50 50 29.6°
60 60 34°
70 70 37.3°
80 80 39.4°
90 90 40.2°

Explanation of Solution

From, law of reflection, the angle of reflection is equal to the angle of incidence. So for all value of angles between 0° to 90° the angle of incidence and angle of reflection are equal.

The formula to calculate the angle of refraction is,

From, Snell’s law,

n1sinθ1=n2sinθ2 (3)

Substitute 1 for n1 , 1.55 for n2 , 10° for θ1 , in equation (3) to calculate θ2 .

1×sin10°=1.55sinθ2sinθ2=sin10°1.55θ2=sin1(0.17361.55)=6.43° ,

Substitute 1 for n1 , and 1.55 for n2 , 20° for θ1 in equation (3) to calculate θ2 .

1×sin20°=1.55sinθ2sinθ2=sin20°1.55θ2=sin1(0.3421.55)=12.74° .

The remaining values of the angle of refraction can be calculated by the same method.

The table for angle of reflection and angle of refraction for all angles of incidence at 10.0° intervals from 0° to 90° for rays incident from air onto the air-glass surface is,

Angle of incidence Angle of reflection Angle of refraction
0 0 0
10 10 6.43°
20 20 12.74°
30 30 18.81°
40 40 24.5°
50 50 29.6°
60 60 34°
70 70 37.3°
80 80 39.4°
90 90 40.2°

Conclusion:

Therefore, the angle of reflection is same as the angle of incidence and the of angle of refraction for rays incident from air onto the air-glass surface is increasing as angle of incidence is increasing.

The table for angle of reflection and angle of refraction for all angles of incidence at 10.0° intervals from 0° to 90° for rays incident from air onto the air-glass surface is,

Angle of incidence Angle of reflection Angle of refraction
0 0 0
10 10 6.43°
20 20 12.74°
30 30 18.81°
40 40 24.5°
50 50 29.6°
60 60 34°
70 70 37.3°
80 80 39.4°
90 90 40.2°

(d)

Expert Solution
Check Mark
To determine
The angles of reflection and refraction for all angles of incidence at 10.0° intervals from 0° to 90° in a tabular form for rays coming up to the interface through the glass.

Answer to Problem 35.52AP

The angle of reflection is same as the angle of incidence and the of angle of refraction for rays coming up to the interface through the glass will increase up to angle of incidence 40° , after that total internal reflection will occur and there will be no refraction.

The table for angle of reflection and angle of refraction for all angles of incidence at 10.0° intervals from 0° to 90° for rays coming up to the interface through the glass is,

Angle of incidence( ° )

( i )

Angle of reflection( ° )

Angle of refraction( ° )

=sin1(1.55×sini)

0 0 0
10 10 15.6°
20 20 32°
30 30 50.8°
40 40 85°
50 50 No Refraction
60 60 No Refraction
70 70 No Refraction
80 80 No Refraction
90 90 No Refraction

Explanation of Solution

From, the law of reflection, angle of reflection is equal to the angle of incidence. So, for all value of angles between 0° to 90° the angle of incidence and angle of reflection are equal.

The formula to calculate the angle of refraction is,

From, Snell’s law,

n1sinθ1=n2sinθ2 (4)

Substitute 1.55 for n1 , and 1 for n2 , 10° for θ1 , in equation (4) to calculate θ2 .

1.55×sin10°=1×sinθ2sinθ2=1.55×0.1736θ2=sin1(1.55×0.1736)=15.6° .

Substitute 1.55 for n1 , and 1 for n2 , 20° for θ1 , in equation (4) to calculate θ2 .

1.55×sin20°=1×sinθ2sinθ2=1.55×0.342θ2=sin1(1.55×0.342)=32° .

The remaining values of the angle of refraction can be calculated by the same method.

The formula to calculate the critical angle is,

θc=sin1(n2n1) (5)

Here,

θc is the critical angle.

n1 is the index of refraction of glass.

n2 is the index of refraction of air.

Substitute 1.55 for n1 , and 1 n2 in equation (5) to calculate θc .

θc=sin1(11.55)=40.2°

The angle of incidence is greater than 40.2° will undergo total internal reflection. Hence, no refraction will occur.

The table for angle of reflection and angle of refraction for all angles of incidence at 10.0° intervals from 0° to 90° for rays coming up to the interface through the glass is,

Angle of incidence( ° )

( i )

Angle of reflection( ° )

Angle of refraction( ° )

=sin1(1.55×sini)

0 0 0
10 10 15.6°
20 20 32°
30 30 50.8°
40 40 85°
50 50 No Refraction
60 60 No Refraction
70 70 No Refraction
80 80 No Refraction
90 90 No Refraction

Conclusion:

Therefore, the angle of reflection is same as the angle of incidence and the of angle of refraction for rays coming up to the interface through the glass will increase up to angle of incidence 40° , after that total internal reflection will occur and there will be no refraction.

The table for angle of reflection and angle of refraction for all angles of incidence at 10.0° intervals from 0° to 90° for rays coming up to the interface through the glass is,

Angle of incidence( ° )

( i )

Angle of reflection( ° )

Angle of refraction( ° )

=sin1(1.55×sini)

0 0 0
10 10 15.6°
20 20 32°
30 30 50.8°
40 40 85°
50 50 No Refraction
60 60 No Refraction
70 70 No Refraction
80 80 No Refraction
90 90 No Refraction

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Chapter 35 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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