Exam 2 practice problems solutions

# 1 1 sin sin 160 sin sin sin sin19 31 1 i i refr refr

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1 1 sin sin 1.60 sin sin sin sin19 31 1 i i refr refr i refr i refr n n n n We could keep ray tracing the reflected beam around the prism, but this is plenty. 23. While you and your friend Sam are studying for your physics exam, Sam asks “This is strange. It seems diverging mirrors can’t produce real images. What do you think?” If you agree with Sam, give as strong an argument as you can. Use as many ray diagrams as you need to support your argument. If you disagree with Sam, give as strong an argument as you can. Use as many ray diagrams as you need to support your argument. I agree with Sam. Qualitatively, all ray diagrams for diverging (convex) mirrors all look the same. The image is always upright, virtual, and demagnified, regardless of the location of the object. Another way to see this is to use the mirror equation. 45 45 135 26 19 71 71 19 31 19 45 45 90 F

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1 1 1 i o d f d Since it’s a diverging mirror, 0 f . Since 0 o d in single mirror situations, the right hand side of the equation will always be negative. This means the images produced by a diverging mirror will always be virtual. 24. You are a professional photographer and use a high quality 16 megapixel digital camera for your work. Inside this camera is a sensor that is used to capture the images. In your camera, this sensor is 3.5 cm in size. The focal length of the camera’s lens is 3.0 cm. You plan on taking pictures of some sculptures that are on exhibit at the loc al art gallery. You want the pictures to be as detailed as possible so you want your camera be close to the sculptures when you take the pictures. For security reasons though, you can’t bring the camera closer than 80 cm. a. Draw a ray diagram for this situation. Show clearly the kind of lens (converging or diverging) that is in your camera, and whether the image is real or virtual. Explain your reasoning. b. Determine the maximum size the sculpture can be so that the camera can still fit it in a single image. In order for the camera to function properly it must create demagnified real images. This requires a converging lens. A side effect of this is that the images will be inverted. In order for the pictures to be as detailed as possible the image must be as large as possible. Since the sensor is 3.5 cm in size, 3.5 cm i h   . It’s negative because the image is inverted. Also, in order for the pictures to be as detailed as possible, the camera lens should be as close to sculpture as possible. This means 80 cm o d . The quantity we are interested in is the height of the object, o h . This quantity only shows up in the magnification equation so we’ll start there.
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