Chapter-34 - 1319 Chapter 34 1. The image is 10 cm behind...

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Unformatted text preview: 1319 Chapter 34 1. The image is 10 cm behind the mirror and you are 30 cm in front of the mirror. You must focus your eyes for a distance of 10 cm + 30 cm = 40 cm. 2. The bird is a distance d 2 in front of the mirror; the plane of its image is that same distance d 2 behind the mirror. The lateral distance between you and the bird is d 3 = 5.00 m. We denote the distance from the camera to the mirror as d 1 , and we construct a right triangle out of d 3 and the distance between the camera and the image plane ( d 1 + d 2 ). Thus, the focus distance is ( 29 ( 29 ( 29 2 2 2 2 1 2 3 4.30m+3.30m 5.00m 9.10m. d d d d = + + = + = 3. When S is barely able to see B the light rays from B must reflect to S off the edge of the mirror. The angle of reflection in this case is 45°, since a line drawn from S to the mirror’s edge makes a 45° angle relative to the wall. By the law of reflection, we find 3.0m tan 45 1 1.5m. / 2 2 2 x d x d = ° = ⇒ = = = 4. The intensity of light from a point source varies as the inverse of the square of the distance from the source. Before the mirror is in place, the intensity at the center of the screen is given by I P = A / d 2 , where A is a constant of proportionality. After the mirror is in place, the light that goes directly to the screen contributes intensity I P , as before. Reflected light also reaches the screen. This light appears to come from the image of the source, a distance d behind the mirror and a distance 3 d from the screen. Its contribution to the intensity at the center of the screen is 2 2 . (3 ) 9 9 P r I A A I d d = = = The total intensity at the center of the screen is 10 . 9 9 P P r P P I I I I I I = + = + = The ratio of the new intensity to the original intensity is I / I P = 10/9 = 1.11. 5. We apply the law of refraction, assuming all angles are in radians: CHAPTER 34 1320 sin sin , θ θ ′ = n n w air which in our case reduces to θ ' ≈ θ / n w (since both θ and θ ' are small, and n air ≈ 1). We refer to our figure on the right. The object O is a vertical distance d 1 above the water, and the water surface is a vertical distance d 2 above the mirror. We are looking for a distance d (treated as a positive number) below the mirror where the image I of the object is formed. In the triangle O AB 1 1 | | tan , AB d d θ θ = ≈ and in the triangle CBD 2 2 2 2 | | 2 tan 2 . w d BC d d n θ θ θ ′ ′ = ≈ ≈ Finally, in the triangle ACI , we have | AI | = d + d 2 . Therefore, ( 29 1 2 2 2 2 2 2 1 2 2 2 | | | | | | 1 | | tan 2 200cm 250cm 200cm 351cm. 1.33 w w d d d AC AB BC d AI d d d d d d n n θ θ θ θ θ   + =- =- ≈- = +- = +-     = +- = 6. The graph in Fig. 34-34 implies that f = 20 cm, which we can plug into Eq. 34-9 (with p = 70 cm) to obtain i = +28 cm....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.

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Chapter-34 - 1319 Chapter 34 1. The image is 10 cm behind...

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