
Concept explainers
(a)
The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and outside air.
(a)

Answer to Problem 28P
Explanation of Solution
Given info: The condition for light ray travelling between air and a diamond is shown below.
Figure (I)
From Snell’s law of refraction to air-diamond interface to find the critical angle is,
Here,
The value of
Substitute 1 for
Thus, the critical angle of refraction at air-diamond interface is
Conclusion:
Therefore, the critical angle of refraction at air-diamond interface is
(b)
To show: The light travelling towards point
(b)

Answer to Problem 28P
Explanation of Solution
Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).
The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and outside air is
Conclusion:
Therefore, the angle of incidence is more than the critical angle all light is reflected from point P.
(c)
The critical angle for total internal reflection for light in the diamond when the diamond is immersed in the water.
(c)

Answer to Problem 28P
Explanation of Solution
Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).
From Snell’s law of refraction to water-diamond interface to find the critical angle is,
Here,
The value of
Substitute 1 for
Thus, the critical angle of incidence at water-diamond interface is
Conclusion:
Therefore, the critical angle of incidence at water-diamond interface is
(d)
The ray incident at point P
undergoes total internal reflection or not when diamond is immersed in the water.
(d)

Answer to Problem 28P
Explanation of Solution
Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).
The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and water is
Thus, the light undergoes total internal reflection at
Conclusion:
Therefore, the light undergoes total internal reflection at
(e)
The direction in which the diamond is rotated such that the light at a point P
will exit the diamond.
(e)

Answer to Problem 28P
Explanation of Solution
Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).
The critical angle for total internal reflection for light in the diamond incident on the interface between the diamond and water is
The light will exit from the diamond only when the incident angle is less than the critical angle. So, to reduce the angle of incidence the diamond should be rotated in clockwise direction on the axis perpendicular to the plane of paper.
Thus, the light will exit at point
Conclusion:
Therefore, the light will exit at point
(f)
The angle of rotation at which the light first exit the diamond at point P
.
(f)

Answer to Problem 28P
Explanation of Solution
Given info: The condition for light ray travelling between air and a diamond is shown in figure (I).
Let the angle is rotated clockwise by
Apply Snell’s law at the water-diamond interface.
The condition of the situation is shown below.
Figure (II)
The angle at the vertex
The requirement is that the angle of incidence
Apply Snell’s law and find angle
Substitute
Thus, the diamond is rotated by
Conclusion:
Therefore, the diamond is rotated by
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Chapter 34 Solutions
PHYSICS:F/SCI.+ENGRS.,V.1
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