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AL physics
發問:
1985mc(41),1985mc(39),1983mc(13),1995mc(17),1985mc(40) Please, explain the answers in detail. Thanks! 1985mc(40), I don't understand why the statement 2 is wrong.
95.17. Answer B. decrease and then increase. When the object is far away from the convex lens, it can be regarded as at infinity. And the real image is at the focus. When it is moved towards the lens, the separation between the object and the real image decreases. It will decrease to a minimum when the object is at 2f, where f is the focal length. While the real image is at 2f on the other side of the lens. Thus, the minimum separation is 4f. You can actually prove it by the lens formula 1/f = 1/v + 1/u. Then, when the object is further moved towards the lens, the separation increases again. When the object is at f, the real image is formed at infinity. 85.41. The answer is E. (3) only All the three incident waves are parallel to each other. So, when it is being reflected by the diverging lens, the interpolation of the reflected wave will meet together at the focal plane. For (1), if you only consider the rays, it is possible. But the wavefront should be circular just like the one in (2) or (3). It is not straight anyway. For (2) and (3), the only difference is one is gathered below F, the other gather above F. When you draw a parallel line to the incident rays which passes through the optical centre, you will find out that (2) doesn't make sense since the wavefront is not perpendicular to the propagation of the 'centre' ray. So, only (3) is true. 85.40. Answer D. (1) only (1) This is a fact. (2) You can refer to http://hk.knowledge.yahoo.com/question/question?qid=7009012401568 Then you will know why the image is not erect. It is erect to the intermediate image only. (3) The objective lens should have a long focal length while the eyepiece should have a short one in order to have a large magnification. 2009-07-12 10:45:19 補充： 85.39. Answer A. (1), (2) and (3) (1). You can derive it from the lens formula 1/f = 1/v + 1/u, moving the lens L to the left means decreasing the value of u, with f being unchanged, when u = f, v is undefined, so, parallel rays are obtained. 2009-07-12 10:45:30 補充： (2). Once again, by lens formula 1/f = 1/v + 1/u, with u being unchanged, f being larger (less converging), v will increase to infinity. (3). Please look at the following link: http://s726.photobucket.com/albums/ww265/physicsworld2010/?action=view¤t=physicsworld01Jul120959.jpg 2009-07-12 10:45:39 補充： 83.13. The answer is B. http://s726.photobucket.com/albums/ww265/physicsworld2010/?action=view¤t=physicsworld02Jul121043.jpg 2009-07-12 10:45:43 補充： You will see from the figure, as the refractive index of the glass is larger than that of air, so the light is being refracted towards the normal. And this makes a lateral displacement. If t increases, d will increase also.
其他解答:
AL physics
發問:
1985mc(41),1985mc(39),1983mc(13),1995mc(17),1985mc(40) Please, explain the answers in detail. Thanks! 1985mc(40), I don't understand why the statement 2 is wrong.
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最佳解答:95.17. Answer B. decrease and then increase. When the object is far away from the convex lens, it can be regarded as at infinity. And the real image is at the focus. When it is moved towards the lens, the separation between the object and the real image decreases. It will decrease to a minimum when the object is at 2f, where f is the focal length. While the real image is at 2f on the other side of the lens. Thus, the minimum separation is 4f. You can actually prove it by the lens formula 1/f = 1/v + 1/u. Then, when the object is further moved towards the lens, the separation increases again. When the object is at f, the real image is formed at infinity. 85.41. The answer is E. (3) only All the three incident waves are parallel to each other. So, when it is being reflected by the diverging lens, the interpolation of the reflected wave will meet together at the focal plane. For (1), if you only consider the rays, it is possible. But the wavefront should be circular just like the one in (2) or (3). It is not straight anyway. For (2) and (3), the only difference is one is gathered below F, the other gather above F. When you draw a parallel line to the incident rays which passes through the optical centre, you will find out that (2) doesn't make sense since the wavefront is not perpendicular to the propagation of the 'centre' ray. So, only (3) is true. 85.40. Answer D. (1) only (1) This is a fact. (2) You can refer to http://hk.knowledge.yahoo.com/question/question?qid=7009012401568 Then you will know why the image is not erect. It is erect to the intermediate image only. (3) The objective lens should have a long focal length while the eyepiece should have a short one in order to have a large magnification. 2009-07-12 10:45:19 補充： 85.39. Answer A. (1), (2) and (3) (1). You can derive it from the lens formula 1/f = 1/v + 1/u, moving the lens L to the left means decreasing the value of u, with f being unchanged, when u = f, v is undefined, so, parallel rays are obtained. 2009-07-12 10:45:30 補充： (2). Once again, by lens formula 1/f = 1/v + 1/u, with u being unchanged, f being larger (less converging), v will increase to infinity. (3). Please look at the following link: http://s726.photobucket.com/albums/ww265/physicsworld2010/?action=view¤t=physicsworld01Jul120959.jpg 2009-07-12 10:45:39 補充： 83.13. The answer is B. http://s726.photobucket.com/albums/ww265/physicsworld2010/?action=view¤t=physicsworld02Jul121043.jpg 2009-07-12 10:45:43 補充： You will see from the figure, as the refractive index of the glass is larger than that of air, so the light is being refracted towards the normal. And this makes a lateral displacement. If t increases, d will increase also.
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