ch26 - CHAPTER 26 THE REFRACTION OF LIGHT LENSES AND...

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CHAPTER 26 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS CONCEPTUAL QUESTIONS ____________________________________________________________________________________________ 1. REASONING AND SOLUTION Since the index of refraction of water is greater than that of air, the ray in Figure 26.1 a is bent toward the normal at the angle q w1 when it enters the water. According to Snell's law (Equation 26.2), the sine of w1 is given by air 1 1 w1 water water sin sin sin n nn θ == (1) where we have taken n air = 1.000 . When a layer of oil is added on top of the water, the angle of refraction at the air/oil interface is oil and, according to Snell's law, we have air 1 1 oil oil oil sin sin sin n (2) But oil is also the angle of incidence at the oil/water interface. At this interface the angle of refraction is w2 and is given by Snell's law as follows: oil oil oil 1 w2 water water oil water sin sin sin sin n n 1 == = (3) where we have substituted Equation (2) for oil sin . According to Equation (1), this result is equal to w1 sin . Therefore, we can conclude that the angle of refraction as the ray enters the water does not change due to the presence of the oil. ____________________________________________________________________________________________ 2. REASONING AND SOLUTION When light travels from a material with refractive index n 1 into a material with refractive index n 2 , the angle of refraction 2 is related to the angle of incidence 1 by Equation 26.2: 11 2 sin 2 = or 1 21 2 sin ( / )sin 1 = . When n 1 < n 2 , the angle of refraction will be less than the angle of incidence. The larger the value of n 2 , the smaller the angle of refraction for the same angle of incidence. The angle of refraction is smallest for slab B; therefore, slab B has the greater index of refraction. ____________________________________________________________________________________________

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1312 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 3. REASONING AND SOLUTION When an observer peers over the edge of a deep empty bowl, he does not see the entire bottom surface, so a small object lying on the bottom is hidden from view. However, when the bowl is filled with water, the object can be seen. When the object is viewed from the edge of the bowl, light rays from the object pass upward through the water. Since n air < n water , the light rays from the object refract away from the normal when they enter air. The refracted rays travel to the observer, as shown in the figure at the right. When the rays entering the air are extended back into the water, they show that the observer sees a virtual image of the object at an apparent depth that is less than the actual depth, as indicated in the drawing. Therefore, the apparent position of the object in the water is in the line of sight of the observer, even though the object could not be seen before the water was added. Imag Object ____________________________________________________________________________________________ 4. SSM REASONING AND SOLUTION Two identical containers, one filled with water ( n = 1.33) and the other filled with ethyl alcohol ( n = 1.36) are viewed from directly above.
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This note was uploaded on 05/09/2008 for the course PHYS 25 taught by Professor Holland during the Spring '08 term at Pacific.

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ch26 - CHAPTER 26 THE REFRACTION OF LIGHT LENSES AND...

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