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Hwksolns7_Publishers - 23.60. Model: Use the ray model of...

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23.60. Model: Use the ray model of light. The surface is a spherically refracting surface. Visualize: Solve: Because the rays are parallel, s = . The rays come to focus on the rear surface of the sphere, so s = 2 R , where R is the radius of curvature of the sphere. Using Equation 23.21, 12 21 11 2.00 2 nn nn n n n ss R R R −− += ⇒+ = ⇒=
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23.69. Model: Assume the lens is a thin lens and the thin-lens formula applies. Solve: Because we want to form an image of the spider on the wall, the image is real and we need a converging lens. That is, both s and s are positive. This also implies that the spider’s image is inverted, so 1 2 Ms s =− . Using the thin-lens formula with 1 2 , s s ′ = 1 2 11 1 1 1 1 s sf ssf +=⇒+ = 31 3 s f sf ⇒= ⇒= We also know that the spider is 2.0 m from the wall, so s + s = 2.0 m = s + 1 2 s ( ) 1 3 4.0 m 133.3 cm s == Thus, 1 3 44 cm fs and 2.0 m 1.33 m 0.67 m 67 cm. s = = We need a 44 cm focal length lens placed 67 cm from the wall.
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23.73. Visualize: The lens must be a converging lens for this scenario to happen, so we expect f to be positive. In the first case the upright image is virtual 2 (0 ) s ′ < and the object must be closer to the lens than the focal point. The lens is then moved backward past the focal point and the image becomes real 2 ) . s ′ > 11 1 ss f s sf s s +=⇒= + We are given 1 10cm s = and 1 2. m = Solve: Since the first image is virtual, 0. s ′ < We are told the first magnification is 11 1 1 22 0 c m . ms s s ′′ == ⇒ = We can now find the focal length of the lens. (10 cm)( 20 cm) 20 cm 10 cm 20 cm ss f = +− After the lens is moved, 2 2. s =− =− Start with the thin lens equation again. 111 s + = Replace 2 s with . 2 s f + = Now solve for 2 . s 2 2 1 () s s f −+ = 2 2 1 s f = Cancel one 2 . s 2 m f = Multiply both sides by 2 . fs 2 2 2 12 1 (20 cm) 30 cm 2 m m ⎛⎞ −− = ⎜⎟ ⎝⎠ The distance the lens moved is 21 30 cm 10 cm 20 cm. −= = Assess: We knew 2 s needed to be bigger than f ; it is, and is in a reasonable range. The final answer for the distance the lens moved also seems reasonable.
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24.1. The picture will be underexposed if the shutter speed remains the same. The problem stated that D is not changed. 2 2 . D I f If f is increased then I is decreased so the picture is underexposed.
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24.2. The answer is C. Rays from all parts of the object hit all parts of the lens, so the whole image still appears, but because the area of the aperture is smaller then the image will be less bright. Contrary to choice a, the image actually gets sharper because the off-axis rays contribute to blurriness and a smaller aperture eliminates many of those rays.
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24.3. If you try to see underwater with no face mask there is little refraction at the water-cornea boundary because the indices of refraction are so similar. To make up for this loss of refraction (as compared to seeing in air) it would be helpful to have glasses with converging lenses.
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24.5.
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This note was uploaded on 12/26/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

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Hwksolns7_Publishers - 23.60. Model: Use the ray model of...

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