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1 Quiz 3 solutions are in order of the questions for version A. (1). Correct answer C In this situation (R > p > R/2) , the mirror will from a real image with q > p; hence the image is behind you as you look towards the mirror and you cannot see it. (2). Correct answer is D We have to solve 1 2 1 + = 50 q 60 which yields a positive q (real image) which must therefore be inverted (3). Correct answer is E Solving for q in the previous formula gives q = 75cm (4). Correct answer is A If we assume that the image is real, we can use q = 60 and find f = 20. We can then check that this is consistent with the rest of the information given, since this predicts that q =100 if p =25. The other possibility, that the first image is virtual and has q=-60 would yield a focal length which disagrees with the second situation. (5). Correct answer is C In order to produce a parallel beam to the right, the object for the second lens must be at p=f, namely 150 cm to the left of the lens. This position is 60cm to the left of the first lens, which means that the first lens must have made a (virtual) image at q = -60. We can then determine the focal length via 1 1 1 + = 72 −60 f1 (6). Correct answer is C This can be found from the formula ∆y = Lλ d where ∆y is the distance between bright lines. 2 (7). Correct answer is B We just have δλ 1 = λ Nm which gives N = 1/3(475/.48) ' 330 (8). Correct answer is A This is a case where there is an extra phase shift and hence the first constructive interference will happen at a thickness of 1/4 of the wavelength in the soap; d =.25(660/1.33) giving result A (9). Correct answer is B For objects at infinity, the image distance is 50mm; for an object at p=800 mm, the image distance is easily found to be 53.333; hence the distance must vary by 3.33 mm (10). Correct answer is B The angular resolution is 1.44 λ / .85. This must be roughly equal to .03/ L where L is the distance to the object. This leads to answer B