Practice Problem - Chapter 26, Part 2

48. REASONING Since we are given the focal length and the object distance, we can use the thin-lens equation to calculate the image distance. From the algebraic sign of the image distance we can tell if the image is real (image distance is positive) or virtual (image distance is negative). Knowing the image distance and the object distance will enable us to use the magnification equation to determine the height of the image. SOLUTION a. Using the thin-lens equation to obtain the image distance d i from the focal length f and the object distance d o , we find 1 i io 111 1 1 0.00491 cm or 204 cm 88.00 cm 155.0 cm d dfd ? =? = ? = = b. The fact that the image distance is positive indicates that the image is real . c. The magnification equation indicates that the magnification m is ii oo hd m hd ==? where h o and h i are the object height and image height, respectively. Solving for the image height gives () i io o 204 cm 13.0 cm 17.1 cm 155.0 cm d hh d g167g183 g167g183 =? = ? =? g168g184 g168g184 g168g184 g169g185 g169g185 The negative value indicates that the image is inverted with respect to the object. 49. REASONING AND SOLUTION a. According to the thin-lens equation, we have io 111 1 1 25 cm 38 cmdfd =? = ? ? or d i = 15 cm? b. The image is virtual since the image distance is negative. ________________________________________________________________________ ______ 50. REASONING AND SOLUTION The image distance for the first case is 3 io 111 1 1 200.0 mm 3.5 10 mmdfd =? = ? × or d i = 212 mm and, similarly, for the second case it is d i = 201 mm. Thus, the lens must be capable of moving through a distance of 212 mm ? 201 mm = 11 mm or 0.011 m . ________________________________________________________________________ ______ 60. REASONING The thin-lens equation can be used to find the image distance of the first image (the image produced by the first lens). This image, in turn, acts as the object for the second lens. The thin-lens equation can be used again to determine the image distance for the final image (the image produced by the second lens). SOLUTION For the first lens, the object and image distances, d o,1 and d i,1 , are related to the focal length f of the lens by the thin-lens equation o1 i1 111 ddf += (26.6) Solving this expression for the image distance produced by the first lens, we find that 1 i1 o1 111 1 1 or 18.0 cm 12.00 cm 36.00 cm i d dfd =? = ? = This image distance indicates that the first image lies between the lenses. Since the lenses are separated by 24.00 cm, the distance between the image produced by the first lens and the second lens is 24.00 cm ? 18.0 cm = 6.0 cm. Since the image produced by the first lens acts as the object for the second lens, we have that d o2 = 6.0 cm. Applying the thin-lens equation to the second lens gives 2 i2 o2 111 1 1 or 12 cm 12.00 cm 6.0 cm i d dfd =? = ? =? The fact that this image distance is negative means that the final image is virtual and lies to the left of the second lens. ________________________________________________________________________ ______ 70. REASONING The closest she can read the magazine is when the image formed by the contact lens is at the near point of her eye, or d i = ?138 cm. The image distance is negative because the image is a virtual image (see Section 26.10). Since the focal length is also known, the object distance can be found from the thin-lens equation. SOLUTION The object distance d o is related to the focal length f and the image distance d i by the thin-lens equation: o oi 111 1 1 or 28.0 cm 35.1 cm 138 cm d dfd =? = ? = ? (26.6) ________________________________________________________________________ ______ 79. REASONING The angular magnification M of a magnifying glass is given by Equation 26.10 as i 1 MN fd ? ? g167g183? 1 =? ? g168g184 g168g184 g169g185 where ? ? = 0.0380 rad is the angular size of the final image produced by the magnifying glass, ? = 0.0150 rad is the reference angular size of the object seen at the near point without the magnifying glass, and N is the near point of the eye. The largest possible angular magnification occurs when the image is at the near point of the eye, or d i = ?N, where the minus sign denotes that the image lies on the left side of the lens (the same side as the object). This equation can be solved to find the focal length of the magnifying glass. SOLUTION Letting d i = ?N, and solving Equation 26.10 for the focal length f gives 21.0 cm 13.7 cm 0.0380 rad 11 0.0150 rad N f ? ? == = ? ?? ________________________________________________________________________ ______ mrw Microsoft Word - prob_048.doc