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Simple Nature - Light and Matter

Simple Nature - Light and Matter

Simple Nature - Light and Matter

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Page 798, problem 19: (a) The object distance is less than the focal length, so the image isvirtual: because the object is so close, the cone of rays is diverging too strongly for the mirror tobring it back to a focus. (b) At an object distance of 30 cm, it’s clearly going to be real. Withthe object distance of 20 cm, we’re right at the crossing-point between real <strong>and</strong> virtual. Forthis object position, the reflected rays will be parallel. We could consider this to be an imageat infinity. (c),(d) A diverging mirror can only make virtual images.Page 802, problem 39: Since d o is much greater than d i , the lens-film distance d i is essentiallythe same as f. (a) Splitting the triangle inside the camera into two right triangles,straightforward trigonometry givesθ = 2 tan −1 w 2ffor the field of view. This comes out to be 39 ◦ <strong>and</strong> 64 ◦ for the two lenses. (b) For small angles,the tangent is approximately the same as the angle itself, provided we measure everything inradians. The equation above then simplifies toθ = w fThe results for the two lenses are .70 rad = 40 ◦ , <strong>and</strong> 1.25 rad = 72 ◦ . This is a decent approximation.(c) With the 28-mm lens, which is closer to the film, the entire field of view we had with the50-mm lens is now confined to a small part of the film. Using our small-angle approximationθ = w/f, the amount of light contained within the same angular width θ is now striking a pieceof the film whose linear dimensions are smaller by the ratio 28/50. Area depends on the squareof the linear dimensions, so all other things being equal, the film would now be overexposed bya factor of (50/28) 2 = 3.2. To compensate, we need to shorten the exposure by a factor of 3.2.Page 810, problem 59: One surface is curved outward <strong>and</strong> one inward. Therefore the minussign applies in the lensmaker’s equation. Since the radii of curvature are equal, the quantity1/r 1 − 1/r 2 equals zero, <strong>and</strong> the resulting focal length is infinite. A big focal length indicates aweak lens. An infinite focal length tells us that the lens is infinitely weak — it doesn’t focus ordefocus rays at all.940 Chapter Appendix 4: Hints <strong>and</strong> Solutions

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