PHYS
Chapter 23

# Angles but rays from the marks near the top of the

• Notes
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angles, but rays from the marks near the top of the tank are incident at very large angles—greater than the critical angle. These rays undergo total internal reflection in the water and do not exit into the air where they can be seen. Thus you can see the marks from the bottom of the tank upward. (b) The highest point you can see is the one from which the ray reaches the top surface at the critical angle θ c . For a water-air boundary, the critical angle is θ c = sin –1 (1/1.33) = 48.75°. You can see from the figure that the depth of this point is such that c c 65.0 cm tan 57.0 cm tan tan(48.75 ) L L d d θ θ = = = = ° Since the marks are every 10 cm, the high mark you can see is the one at 60 cm.

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23.50. Model: Use the ray model of light and the law of refraction. Assume the sun is a point source of light. Visualize: When the bottom of the pool becomes completely shaded, a ray of light that is incident at the top edge of the swimming pool does not reach the bottom of the pool after refraction. Solve: The depth of the swimming pool is water 4.0 m tan . d θ = We will find the angle by using Snell’s law. We have water water air sin sin70 n n θ = ° ⇒ 1 water sin70 sin 44.95 1.33 θ ° = = ° 4.0 m 4.0 m tan44.95 d = = °
23.51. Model: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of light. Visualize: The laser beam enters the water 2.0 m from the edge, undergoes refraction, and illuminates the goggles. The ray of light from the goggles then retraces its path and enters your eyes. Solve: From the geometry of the diagram, 1.0 m tan 2.0 m φ = ( ) 1 tan 0.50 26.57 φ = = ° air 90 26.57 63.43 θ = ° − ° = ° Snell’s law at the air-water boundary is air air water water sin sin . n n θ θ = Using the above result, ( ) water 1.0 sin63.43 1.33sin θ ° = 1 water sin63.43 sin 42.26 1.33 θ ° = = ° Taking advantage of the geometry in the diagram again, ( ) water tan 3.0 m tan42.26 2.73 m 3.0 m x x θ = = ° = The distance of the goggles from the edge of the pool is 2.73 m + 2.0 m = 4.73 m 4.7 m.

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23.52. Model: Use the ray model of light and the law of refraction. Assume that the laser beam is a ray of light. Visualize: Solve: (a) From the geometry of the diagram at side A, 10 cm tan 15 cm φ = 1 10 tan 15 φ = 33.69 φ = ° This means the angle of incidence at side A is θ air = 90 ° 33.69 ° = 56.31 ° . Using Snell’s law at side A, air air water water A sin sin n n θ θ = 1 water A 1.0sin56.31 sin 38.73 1.330 θ ° = = ° This ray of light now strikes side B. The angle of incidence at this water-air boundary is water B water A 90 51.3 . θ θ = ° − = ° The critical angle for the water-air boundary is 1 1 air c water 1.0 sin sin 48.8 1.33 n n θ = = = ° Because the angle θ water B is larger than θ c , the ray will experience total internal reflection.
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• Winter '10
• E.Salik
• Light, refractive index, Total internal reflection, Geometrical optics

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