Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 38, Problem 9PQ
To determine
The angle of refraction in water.
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Chapter 38 Solutions
Physics for Scientists and Engineers: Foundations and Connections
Ch. 38.1 - Light travels from air into glass. Which sketch in...Ch. 38.2 - Prob. 38.2CECh. 38.3 - Prob. 38.3CECh. 38.6 - Prob. 38.4CECh. 38.7 - Prob. 38.5CECh. 38.9 - Prob. 38.6CECh. 38.9 - Prob. 38.7CECh. 38.10 - Prob. 38.8CECh. 38 - The Sun appears at an angle of 53.0 above the...Ch. 38 - Prob. 2PQ
Ch. 38 - Prob. 3PQCh. 38 - A light ray is incident on an interface between...Ch. 38 - Prob. 5PQCh. 38 - Prob. 6PQCh. 38 - Prob. 7PQCh. 38 - A ray of light enters a liquid from air. If the...Ch. 38 - Prob. 9PQCh. 38 - Figure P38.10 on the next page shows a...Ch. 38 - Prob. 11PQCh. 38 - Prob. 12PQCh. 38 - Prob. 13PQCh. 38 - Prob. 14PQCh. 38 - Prob. 15PQCh. 38 - A fish is 3.25 m below the surface of still water...Ch. 38 - N A fish is 3.25 m below the surface of still...Ch. 38 - A beam of monochromatic light within a fiber optic...Ch. 38 - Prob. 19PQCh. 38 - Prob. 20PQCh. 38 - Consider a light ray that enters a pane of glass...Ch. 38 - Prob. 22PQCh. 38 - Prob. 23PQCh. 38 - Prob. 24PQCh. 38 - Prob. 25PQCh. 38 - Prob. 26PQCh. 38 - Prob. 27PQCh. 38 - Prob. 28PQCh. 38 - The wavelength of light changes when it passes...Ch. 38 - Prob. 30PQCh. 38 - Light is incident on a prism as shown in Figure...Ch. 38 - Prob. 32PQCh. 38 - Prob. 33PQCh. 38 - Prob. 34PQCh. 38 - Prob. 35PQCh. 38 - Prob. 36PQCh. 38 - Prob. 37PQCh. 38 - A Lucite slab (n = 1.485) 5.00 cm in thickness...Ch. 38 - Prob. 39PQCh. 38 - Prob. 40PQCh. 38 - The end of a solid glass rod of refractive index...Ch. 38 - Prob. 42PQCh. 38 - Figure P38.43 shows a concave meniscus lens. If...Ch. 38 - Show that the magnification of a thin lens is...Ch. 38 - Prob. 45PQCh. 38 - Prob. 46PQCh. 38 - Prob. 47PQCh. 38 - The radius of curvature of the left-hand face of a...Ch. 38 - Prob. 49PQCh. 38 - Prob. 50PQCh. 38 - Prob. 51PQCh. 38 - Prob. 52PQCh. 38 - Prob. 53PQCh. 38 - Prob. 54PQCh. 38 - Prob. 55PQCh. 38 - Prob. 56PQCh. 38 - Prob. 57PQCh. 38 - Prob. 58PQCh. 38 - Prob. 59PQCh. 38 - Prob. 60PQCh. 38 - Prob. 61PQCh. 38 - Prob. 62PQCh. 38 - Prob. 63PQCh. 38 - Prob. 64PQCh. 38 - Prob. 65PQCh. 38 - Prob. 66PQCh. 38 - Prob. 67PQCh. 38 - Prob. 68PQCh. 38 - CASE STUDY Susan wears corrective lenses. The...Ch. 38 - A Fill in the missing entries in Table P38.70....Ch. 38 - Prob. 71PQCh. 38 - Prob. 72PQCh. 38 - Prob. 73PQCh. 38 - Prob. 74PQCh. 38 - An object 2.50 cm tall is 15.0 cm in front of a...Ch. 38 - Figure P38.76 shows an object placed a distance...Ch. 38 - Prob. 77PQCh. 38 - Prob. 78PQCh. 38 - Prob. 79PQCh. 38 - CASE STUDY A group of students is given two...Ch. 38 - A group of students is given two converging...Ch. 38 - Prob. 82PQCh. 38 - Two lenses are placed along the x axis, with a...Ch. 38 - Prob. 84PQCh. 38 - Prob. 85PQCh. 38 - Prob. 86PQCh. 38 - Prob. 87PQCh. 38 - Prob. 88PQCh. 38 - Prob. 89PQCh. 38 - Prob. 90PQCh. 38 - Prob. 91PQCh. 38 - Prob. 92PQCh. 38 - Prob. 93PQCh. 38 - Prob. 94PQCh. 38 - Prob. 95PQCh. 38 - Prob. 96PQCh. 38 - Prob. 97PQCh. 38 - A Fermats principle of least time for refraction....Ch. 38 - Prob. 99PQCh. 38 - Prob. 100PQCh. 38 - Prob. 101PQCh. 38 - Prob. 102PQCh. 38 - Prob. 103PQCh. 38 - Prob. 104PQCh. 38 - Curved glassair interfaces like those observed in...Ch. 38 - Prob. 106PQCh. 38 - Prob. 107PQCh. 38 - Prob. 108PQCh. 38 - Prob. 109PQCh. 38 - Prob. 110PQCh. 38 - Prob. 111PQCh. 38 - Prob. 112PQCh. 38 - Prob. 113PQCh. 38 - Prob. 114PQCh. 38 - The magnification of an upright image that is 34.0...Ch. 38 - Prob. 116PQCh. 38 - Prob. 117PQCh. 38 - Prob. 118PQCh. 38 - Prob. 119PQCh. 38 - Prob. 120PQCh. 38 - Prob. 121PQCh. 38 - Prob. 122PQCh. 38 - Prob. 123PQCh. 38 - Prob. 124PQCh. 38 - Prob. 125PQCh. 38 - Prob. 126PQCh. 38 - Light enters a prism of crown glass and refracts...Ch. 38 - Prob. 128PQCh. 38 - An object is placed a distance of 10.0 cm to the...
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- A block of crown glass is immersed in water as in the figure below. A light ray is incident on the top face at an angle of θ1= 41° with the normal and exits the block at point P. Find the angle of refraction θ2 of the light ray leaving the block at P. 80.2° 41° 43.3° 68.9°arrow_forwardFigure P23.28 shows a curved surface separating a material with index of refraction n1 from a material with index n2 . The surface forms an image I of object o. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to n1 sin θ1 = n2 sin02 . For paraxial rays, we assume θ1 and θ2 are small, so we may write n1 tan θ1 = n2 tan θ2. The magnification is defined as M =h′/h. Prove that the magnification is given by M = −n1 q/n2p.arrow_forwardTwo plane mirrors are facing each other. They are parallel, 3.00 cm apart, and 17.0 cm in length, as the drawing indicates. A laser beam is directed at the top mirror from the left edge of the bottom mirror. What is the smallest angle of incidence with respect to the top mirror, such that the laser beam (a) hits only one of the mirrors and (b) hits each mirror only once? Laser (a) Number (b) Number i 17.0 cm Units Units 마음이 3.00 cm >arrow_forward
- Choose the correct statement regarding light traveling in air and glass mediums. Assume that the angle of incidence is not perpendicular to the surface. Refractive index of air is nair=1.00029; refractive index of glass is nglass=1.517. For light traveling from glass to air, the ray becomes bent toward the normal. O Light travels at a slower speed in air than in glass. For light traveling from air to glass, the ray becomes bent away from the normal. O For light traveling from air to glass, the incidence angle is larger than the refraction angle. O For light traveling from glass to air, the refraction angle is smaller than the incidence angle. Submit Answer Tries 0/2 Post Discussion Send Feedbaclarrow_forwardAs shown in the figure, a ray of light strikes a plane mirror with some incident angle. The mirror is now rotated by an angle of ? = 19.0° about an axis through the point where N1 contacts the mirror, without altering the incident ray. The new position is shown by the line M2. (a) Determine the angle through which the reflected ray rotates if the incident angle is 40.0°.° (b) Determine the angle through which the reflected ray rotates if the incident angle is 50.0°.°arrow_forwardThe drawing shows a laser beam shining on a plane mirror that is perpendicular to the floor. The angle of incidence is 33.0°. The beam emerges from the laser at a point that is 1.10 m from the mirror and 1.80 m above the floor. After reflection, how far from the base of the mirror does the beam strike the floor?arrow_forward
- A plane mirror and a concave mirror (f = 7.80 cm) are facing each other and are separated by a distance of 23.5 cm. An object is placed between the mirrors and is 11.8 cm from each mirror. Consider the light from the object that reflects first from the plane mirror and then from the concave mirror. What is the distance of the image (di) produced by the concave mirror?arrow_forwardWater floats on a liquid called carbon tetrachloride. The two liquids do not mix. A light ray passing from water into carbon tetrachloride has an incident angle of 45.0° and an angle of refraction of 40.1°. If the index of refraction of water is 1.33, what is the index of refraction of carbon tetrachloride?arrow_forwardConsider a ray incident on the end of a fiber-optic cable as illustrated in the figure below. In the experiment, you will explore the behavior of the fiber as a function of the angle of incidence a. When a ray strikes the end of the fiber it is bent toward the normal. By Snell's law, the angle inside the core is sin core = (no/n₁) sin a. Once inside the core, the ray travels until it strikes the cladding of the fiber. The angle of incidence at the cladding i is the complementary angle of core. If i is a small angle, the ray will propagate into the cladding and be lost from the fiber. If i is large, however, the ray will be internally reflected and bounce down the fiber. By Snell's law, the critical angle for the ray to be internally reflected is sin i crit n₂In₁. In the experiment, you cannot directly measure angles inside the fiber but you can measure a. Using the facts that, (i) for complementary angles, √T sin² i, and (iii) the index of refraction of air is no = 1, it can be shown…arrow_forward
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