Exam 2 practice problems solutions

# B are the objects actually closer to the mirror than

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b. Are the objects actually closer to the mirror than their images? Explain your reasoning. c. If the objects are actually further away from the mirror than their images, then why does it look like the images are more distant? The mirror must produce upright, virtual images for all objects. Diverging (convex) mirrors accomplish this. The objects are not closer to the mirror than their images since for diverging mirrors i o d d . The reason the images look distant is not because they actually are further away than the actual objects, but because they are demagnified significantly. 21. Make as strong an argument as you can to disprove the following hypothesis: “It is possible for a single diverging lens to produ ce a real image.” Include whatever diagrams, calculations, and numerical values you need to make your argument. F F F

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Lenses (and mirrors also) have three regions where the object can placed that could potentially produce in images with distinctly different characteristics. 1) Between the focal point of the lens and the vertex of the lens, 2) Between the focal point and the center of the lens, and 3) Beyond the center of the lens. Drawing a ray diagram for each case shows that a diverging lens can only produce virtual images. 22. A beam of white light traveling in air (index of refraction 1.00) enters a glass prism (average index of refraction 1.60) as shown. Ray trace what happens to the beam as it interacts with the prism. Label all relevant angles with numerical values. Show your work. 45 F C V F C V F C V
A lot happens to the beam. F irst, reflection and refraction happen. The refraction angle is determined by Snell’s law. 1 1 sin sin 1 sin sin sin sin 45 26 1.60 i i refr refr i refr i refr n n n n Dispersion will also happen, splitting the white light beam into its component colors with the short wavelengths (violet) bending a bit more and the long wavelengths (red) bending a bit less. We don’t have specific information about how the index of refraction of the prism depends on wavelength, so we’ll just use the average value and not ray trace the individual colors from here on. The next thing that happens is the beam interacts with the bottom face of the prism. Total internal reflection could occur. To decide if it does we’ll determine the critical angle. 1 1 sin sin 1 sin sin sin sin90 39 1.60 i i refr refr refr i refr i n n n n Since the incident angle is greater than this, total internal reflection occurs. The last thing that happens is the beam is incident on the right face of the prism. Since the incident angle is less than the critical angle both reflection and refraction occur. Snell’s law gives us the refraction angle.

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