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Physics 112 Exam 3 Summer 2010 41. Suppose that you want to take a photograph of yourself as you look at your image in a flat mirror 2.5 m away. For what distance should the camera lens be focused? A) 2.5m B) 5.0m C) 6.25 m D) 25.0 m E) none of these Solution: For a flat mirror the image is as far behind the mirror as the object is in front, so the distance from object to image is do di 2.5m 2.5m 5.0m. 42. The speed of light in a certain substance is 89% of its value in water. What is the index of refraction of this substance? (The index of refraction of water is 1.33) A) 1.00 B) 1.18 C) 1.33 D) 1.49 E) 1.52 Solution: We find the index of refraction from n n c c 1.33 water n 1.49 v 0.89v water 0.89 0.89 43. A ray of light is refracted through three different materials. Rank the materials according to their index of refraction. A) n1 n2 n3 B) n1 n3 n2 C) n2 n1 n3 D) n2 n3 n1 E) n3 n2 n1 Solution: By looking at the direction and the relative amount that the light rays bend at each interface, we can infer the relative sizes of the indices of refraction in the different materials (bends toward normal = smaller n material to larger n material; bends away from normal = larger n material to smaller n material). From the first material to the second material the ray bends toward the normal, thus n1 < n2. From the second material to the third material the ray bends away from the normal, thus n2 > n3. Careful inspection shows that the ray in the third material does not bend back away from the normal as far as the ray was in the first material, thus and n1 < n3. Thus, the overall ranking of indices of refraction is: n1 < n3 < n2. Physics 112 Exam 3 Summer 2010 44. A substance has an index of refraction of 1.46. Light is passing through it at 53.0°. At what angle will it leave into the air? A) It will not leave. B) 59.1° C) 53.0° D) 43.2° E) 33.2° Solution: The critical angle for the total internal refractions is given by the equation: c 43.2 53.0 . sin c n2 n1 sin c 1 1.46 45. Is it possible to see a virtual image? A) No, since the rays that seem to emanate from a virtual image do not in fact emanate from the image. B) No, since virtual images do not really exist. C) Yes, the rays that appear to emanate from a virtual image can be focused on the retina just like those from an illuminated object. D) Yes, since almost everything we see is virtual because most things do not themselves give off light, but only reflect light coming from some other source. E) Yes, but only indirectly in the sense that if the virtual image is formed on a sheet of photographic film, one could later look at the picture formed. Solution: Yes, the rays coming from the real object can appear to emanate from a virtual image, and can be focused on the retina. 46. Where should an object be placed in front of a concave mirror so that it produces an image at the same location as the object? Is the image real or virtual? Is the image inverted or upright? A) At the focus. It is real and inverted. B) At the focus. It is real and upright. C) At the focus. It is virtual and inverted. D) At the center of curvature. It is real and inverted. E) At the center of curvature. It is virtual and inverted. Solution: a) With di do , we locate the object from 1 1 1 ; d o di f Physics 112 Exam 3 Summer 2010 1 1 1 , which gives d o 2 f r. do do f The object should be placed at the center of curvature. b) Because the image is in front of the mirror, d i 0, it is real. c) The magnification is m di d o 1. do do Because the magnification is negative, the image is inverted. 47. A single concave spherical mirror produces an image which is A) always virtual B) always real C) real only if the object distance is less than f D) real only if the object distance is greater than f E) real only if the object distance is greater than 2f 48. An object is placed between a converging lens and its focal point. The image formed is A) virtual and erect B) virtual and inverted C) real and erect D) real and inverted E) image is not formed 49. A glass converging lens has one flat side and another with a radius of 10 cm. What is the focal length of the lens? ( The index of refraction of the glass in the lens is 1.50) A) 10 cm B) 20 cm C) 30 cm D) 40 cm E) 50 cm Solution: R1 R2 10cm n 1.5 f ? 1 1 1 n 1 f R R 2 1 1 1 1 1.5 1 0 f 10cm 20cm f 20cm Physics 112 Exam 3 Summer 2010 50. An object is placed in front of converging lens with focal distance 15 cm. What is the distance between the object and the lens if magnification is 3? A) 5 cm B) 10 cm C) 15 cm D) 20 cm E) 25 cm Solution: d m i d i md 0 d0 1 1 1 d0 di f 1 d 0 f 1 ; m 1 1 1 d 0 md 0 f 1 1 1 1 d0 m f 1 d 0 15cm 1 3 d 0 10cm . 51. In a Young's double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is 2.5 m, find the spacing between the first-order and second-order bright fringes for light with wavelength of 600 nm. A) 1.0 cm B) 2.0 cm C) 3.0 cm D) 4.0 cm E) 5.0 cm Solution: Given: d 0.050mm 600nm x 2.5m y2 y1 ? Conditions for constrictive interference: From geometry: Because / d 1 sin m tan m . Combining these equations we have: y2 y1 x / d y2 y1 3.0cm d sin m m tan m y m x dy m x m y m m x d y 2 y1 600 10 9 m 2.5m / 0.050 10 3 m 3.0 10 2 m Physics 112 Exam 3 Summer 2010 52. The separation between adjacent maxima in a double-slit interference pattern using monochromatic light is A) greatest for red light B) greatest for green light C) greatest for blue light D) the same for all colors of light E) greatest for red light or for blue light depending on the distance between the slits Solution: Conditions for constrictive interference: d sin m m . For given index m, angle m is increasing with increasing wavelength . Maximum value of m is for the maximum value of , which is for red light. 53. In order to obtain a good single slit diffraction pattern, the slit width could be: A) /100 B) /10 C) D) 10 E) 100 Solution: D sin m m m 1,2,3,... If D 10 , than sin m 0.1m and one can observe several dark and bright fringes. 54. Monochromatic light falls on a slit that is 2.20 10 3 mm wide. If the angle between the first dark fringes on either side of the central maximum is 35.0° (dark fringe to dark fringe), what is the wavelength of the light used? A) 466nm B) 552nm C) 585nm D) 610nm E) 662nm Solution: The angle from the central maximum to the first minimum is 1 35 / 2 17.5 . We find the wavelength from D sin 1 D sin m m 2.20 10 mm sin 17.5 , 3 m 1,2,3,... which gives 6.62 10 7 m 662nm Physics 112 Exam 3 Summer 2010 55. What is the highest spectral order that can be seen if a grating with 6000 lines per cm is illuminated with 633-nm laser light? Assume normal incidence. A) 2 B) 3 C) 4 D) 5 E) 6 Solution: d sin m m m 0,1,2,... The maximum angle is 90°, so we have 2 1 10 m / cm m 6000 lines / cm 90 and m d . d sin 90 633 10 m 2.6 . 9 Thus two orders can be seen on each side of the central white line. 56. A lens appears greenish yellow ( 570 nm is strongest) when white light reflects from it. What minimum thickness of coating (n 1.25) do you think is used on such a glass (n 1.52) lens? A) 180nm B) 205nm C) 228nm D) 350nm E) 420nm Solution: There are a phase shifts ( / 2 ) on both boundaries of the coating film. The total phase shift is . Constrictive interference should be observed if 2t m film with minimum m=1. film 570nm t min 228nm 2 2n 2 1.25 Physics 112 Exam 3 Summer 2010 57. A binary star system in the constellation Orion has an angular separation of 10-5 rad. If the wavelength of the light from the system is 500 nm, what is the smallest aperture (diameter) telescope that can just resolve the two stars? A) 0.50 cm B) 0.61 cm C) 5.0 cm D) 6.1 cm E) 7.0 cm Solution: 1 sin D 1.22 sin 1.22 D sin 1.22 9 D 1.22500 10 m 10 5 6.1 10 2 m 6.1cm 58. A magnifying glass with a focal length of 7.5cm is used to read print placed at a distance 5.0 cm. Calculate the angular magnification. A) 1 B) 2 C) 3 D) 4 E) 5 Solution: M ' 25cm 25cm 5 d0 5.0cm 59. A person’s eye is corrected by a 4.00-diopter lens. What is this eye’s far point without glasses? (Assume that the distance between the eye and the lens can be neglected.) A) 20cm B) 25cm C) 30cm D) 35cm E) 40cm Solution: We find the far point by finding the image distance for an object at infinity ( d 0 ): 1 1 1 1 1 P 4.00 D d i 0.250m 25.0cm d0 di f di Physics 112 Exam 3 Summer 2010 60. An astronomical telescope has its two lenses spaced 90.0 cm apart. If the objective lens has a focal length of 89.0 cm, what is the magnification of this telescope? Assume a relaxed eye. A) -45 B) -80 C) -90 D) -110 E) -120 Solution: For both object and image far away, we find the focal length of the eyepiece from the separation of the lenses: L fo fe ; f e L f 0 90.0cm 89.0cm 1.0cm The magnification of the telescope is given by M fo 90.0cm 90 . fe 1.0cm Physics 112 Exam 3 Summer 2010 Record Sheet You may fill in this sheet with your choices, detach it and take it with you after the exam for comparison with the posted answers 41 B) 5.0m 51 C) 3.0 cm 42 D) 1.49 52 A) greatest for red light 43 B) n1 n3 n2 53 D) 10 44 A) It will not leave. 54 E) 662nm 45 C) Yes, the rays that 55 A) 2 appear to emanate from a virtual image can be focused on the retina just like those from an illuminated object. 46 D) At the center of curvature. It is real and inverted. 47 D) real only if the object distance is greater than f 56 C) 228nm 48 A) virtual and erect 58 E) 5 49 B) 20 cm 59 B) 25cm 50 B) 10 cm 60 C) -90 57 D) 6.1 cm