Chapter 25 - CHAPTER 25 Quick Quizzes 1. We would like to...

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C HAPTER 25 Quick Quizzes 1. We would like to reduce the minimum angular separation for two objects below the angle subtended by the two stars in the binary system. We can do that by reducing the wavelength of the light—this in essence makes the aperture larger, relative to the light wavelength, increasing the resolving power. Thus, we would choose a blue filter. 301
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CHAPTER 25 Problem Solutions 25.1 Using the thin lens equation, the image distance is ( )( ) 150 cm 25.0 cm 30.0 cm pf q == = −− so the image is located 30.0 cm beyond the lens . The lateral magnification is q M p =− = 1 5 25.2 The f -number of a camera lens is defined as -num ber focal lengt hdiam eter f = . Therefore, the diameter is 55 mm 1.8 f D f = 31 25.3 Consider rays coming from opposite edges of the object and passing undeviated through the center of the lens as shown at the right. For a very distant object, the image distance equals the focal length of the lens. If the angular width of the object is θ , the full image width on the film is () 20 2t an 2 2 55.0 mm tan 2 hf ° ⎛⎞ ⎡⎤ = 19 ⎜⎟ ⎣⎦ ⎝⎠ so the image easily fits within a 23.5 mm by 35.0 mm area. θ /2 f h (image size) angular width of object = θ 25.4 The image distance is q f since the object is so far away. Therefore, the lateral magnification is Mh hq p f p −≈ and the diameter of the Moon’s image is ( ) 6 8 120 22 1 . 74 10 m 3.84 m oon f hM h R p = × = 1.09 × 302
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CHAPTER 25 25.5 The exposure time is being reduced by a factor of 2 1 1 256 s 1 132 s8 t t = = Thus, to maintain correct exposure, the intensity of the light reaching the film should be increased by a factor of 8. This is done by increasing the area of the aperture by a factor of 8, so in terms of the diameter, ( ) 22 21 48 4 DD π = or 8 = . The new f -number will be () ( ) 1 2 2 1 -num ber 4.0 1.4 88 8 f ff f D D == = or f 25.6 (a) The intensity is a measure of the rate at which energy is received by the film per unit area of the image , or 1 image IA . Consider an object with horizontal and vertical dimensions as shown at the right. If the vertical dimension intercepts angle θ , the vertical dimension of the image is and x h y h y hq = , or y . Similarly for the horizontal dimension, x h q , and the area of the image is 2 x y Ah h q = ′′ . Assuming a very distant object, q f , so and we conclude that 2 A f 2 1 If . The intensity of the light reaching the film is also proportional to the area of the lens, and hence to the square of the diameter of that lens, or . Combining this with our earlier conclusion gives 2 ID 2 2 2 1 D I f fD ∝= or 2 1 I f θ h x h y h y h x θ q f (b) The total light energy hitting the film is proportional to the product of intensity and exposure time, It . Thus, to maintain correct exposure, this product must be kept constant, or giving 11 It = 2 2 2 1 1 2 2 1 4. 01 s 1. 85 0 0 f I tt t I f ⎡⎤ ⎛⎞ ⎢⎥ = ⎜⎟ ⎝⎠ ⎣⎦ 1 s 100 303
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CHAPTER 25 25.7 Since the exposure time is unchanged, the intensity of the light reaching the film should be doubled so the energy delivered will be doubled. Using the result of Problem 6 (part
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Chapter 25 - CHAPTER 25 Quick Quizzes 1. We would like to...

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