# ch34 - Chapter 34 1. The bird is a distance d2 in front of...

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1305Chapter 341. The bird is a distanced2in front of the mirror; the plane of its image is that samedistanced2behind the mirror. The lateral distance between you and the bird isd3= 5.00m. We denote the distance from the camera to the mirror asd1, and we construct a righttriangle out ofd3and the distance between the camera and the image plane (d1+d2).Thus, the focus distance is22221234.30 m3.30 m5.00 m9.10 m.dddd2. The image is 20 cm behind the mirror and you are 30 cm in front of the mirror. Youmust focus your eyes for a distance of 20 cm + 30 cm = 50 cm.3. The intensity of light from a point source varies as the inverse of the square of thedistance from the source. Before the mirror is in place, the intensity at the center of thescreen is given byIP=A/d2, whereAis a constant of proportionality. After the mirror isin place, the light that goes directly to the screen contributes intensityIP, as before.Reflected light also reaches the screen. This light appears to come from the image of thesource, a distancedbehind the mirror and a distance 3dfrom the screen. Its contributionto the intensity at the center of the screen is22.(3 )99PrIAAIddThe total intensity at the center of the screen is10.99PPrPPIIIIIIThe ratio of the new intensity to the original intensity isI/IP= 10/9 = 1.11.4. WhenSis barely able to seeB, the light rays fromBmust reflect toSoff the edge ofthe mirror. The angle of reflection in this case is 45°, since a line drawn fromSto themirror’s edge makes a 45° angle relative to the wall. By the law of reflection, we findxd/ 2tan45 1xd23.5m21.8m.5. We apply the law of refraction, assuming all angles are in radians:
CHAPTER 341306sinsin,nnwairwhich in our case reduces to'/nw(since bothand'are small, andnair1). We refer to our figure on the right.The objectOis a vertical distanced1above the water, andthe water surface is a vertical distanced2above the mirror.We are looking for a distanced(treated as a positivenumber) below the mirror where the imageIof the object isformed. In the triangleO AB11||tan,ABddand in the triangleCBD2222||2tan2.wdBCddnFinally, in the triangleACI, we have |AI| =d + d2. Therefore,22222121222||||||1||tan2 200cm250cm200cm351cm.1.33wwddACABBCdAIdddddddnn6. We note from Fig. 34-34 thatm=12whenp= 5 cm.Thus Eq. 34-7 (the magnificationequation) gives usi=10 cm in that case.Then, by Eq. 34-9 (which applies to mirrorsand thin lenses) we find the focal length of the mirror isf= 10 cm.Next, the problemasks us to considerp= 14 cm.With the focal length value already determined, then Eq.34-9 yieldsi= 35 cm for this new value of object distance.Then, using Eq. 34-7 again,we findm = i/p=2.5.7. We use Eqs. 34-3 and 34-4, and note thatm = –i/p. Thus,1112ppmfr.We solve forp:pr211m35.0cm2112.5010.5 cm.
13078. The graph in Fig. 34-35 implies thatf

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