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1. [-/1 Points] DETAILS SERCP7 22.AE.02. Example 22.2 Angle of Refraction for Glass Goal Apply Snell's law to a slab of glass. Incident Normal Problem A light ray of wavelength 589 nm (produced by a sodium ray lamp) traveling through air is incident on smooth, flat slab of crown glass at an angle of 01 = 33 to the normal, as sketched in Figure 01 22.11. Find the angle of refraction, 02. Strategy Substitute quantities into Snell's law and solve for the unknown angle of refraction, 02. Air Glass Refracted ray Figure 22.11 Refraction of light by glass. Solution Solve Snell's law (Eq. 22.8) for sin 02 sin02 = -sind1 (1) n2 From Table 22.1, find n1 = 1.00 for air and n2 = 1.52 for crown glass. Substitute these values into (1) and sine2 = 1.52 (sin33.00) = 0.358 take the inverse sine of both sides. 02 = sin (0.358) = Remarks Notice the light ray bends toward the normal when it enters a material of a higher index of refraction. If the ray left the material following the same path in reverse, it would bend away from the normal. Exercise 22.2 Hints: Getting Started | I'm Stuck If the light ray moves from inside the glass toward the glass-air interface at an angle of 33.0 to the normal, determine the angle of refraction. The ray bends . away from the normal, as expected.2. [-11 Points] SERCP7 22.AE.04. Example 22.4 Light Passing through a Slab Goal Apply Snell's law when a ray passes into and out of another medium. Problem A light beam traveling through a transparent medium of index of refraction :11 passes through a thick transparent slab with parallel faces and index of refraction in (Fig. 22.12). Show that the emerging beam is parallel to the incident beam. Strategy Apply Snell's law twice, once at the upper surface and once at the lower surface. The two equations will be related because the angle of refraction at the upper surface equals the angle of incidence at the lower surface. The ray passing through the slab makes equal angles with the normal at the entry and exit points. This procedure will enable us to compare angles 91 and 63. Figure 22.12 When light passes through a flat slab of material, the emerging beam is parallel to the incident beam, and therefore 91 : 63. Solution Apply Snell's law to the upper surface. 311162 2 sinl (l) \".2 Apply Snell's law to the lower surface. sinlg = \"2.4an (2) ill Substitute Equation 1 into Equation 2. . :12 n1 _ 51113;; = slli6'1 m 122 Take the inverse sine of both sides, noting that the 93 91 angles are positive and less than 90 Remarks The preceding results prove that the slab doesn't alter the direction of the beam. It does, however, produce a lateral displacement of the beam, as shown in Figure 22.12. Hints: Getting Started | I'm Stuck Suppose the ray, in air with n = 1.00, enters a slab with n = 2.19 at a 393 angle with respect to the normal, then exits the bottom of the slab into water, with n = 1.33. At what angle to the normal does the ray leave the slab? 3. [-11 Points] DETAILS SERCP7 22.AE.06. Example 22.6 A View from the Fish's Eye Goal Apply the concept of total internal reflection. Problem (3) Find the critical angle for a watereair boundary if the index of refraction of water is 1.33. (b) Use the result of part (a) to predict what a sh will see (Fig 22.28) if it looks up toward the water surface at angles of 42.4, 48.8, and 556. Strategy After nding the critical angle by substitution, use the fact that the path of a light ray is reversible: at a given angle, wherever a light beam can go is also where a beam of light can come from, along the same path. Figure 22.28 A sh looks upward toward the water's surface. Solution (a) Find the critical angle for a watereair boundary. Substitute into Equation 22.9 to find the critical \"-1 7 1.00 angle. 111 1.33 (b) Predict what the fish will see if it looks up toward the surface at angles of 42, 49, and 56. 424 I A beam of light shone toward the surface will be completely reflected down toward the bottom of the pool. Reversing the path, the fish sees a reection of 43.80 some object on the bottom. II A beam of light from undenNater will be refracted 55.6 at the surface and enter the air above. Because the path of a light ray is reversible (Snell's law works both going and coming), light from above can follow the same path and be perceived by the fish. III A beam of light will travel through the airewater boundary without changing direction. The sh will see the object exactly as it would in the air. IV Light from underwater is bent so that it travels along the surface. This means that light following the same path in reverse can reach the fish only by skimming along the water surface before being refracted towards the fish's eye. Hints: Getting Started | I'm Stuck Suppose a layer of oil with n = 1.45 coats the surface of the water. What is the critical angle for total internal reflection for light traveling in the oil layer and encountering the oilwater boundary? 6. [-11 Points] SERCP7 22.P.009. A laser beam is incident at an angle of 44.0 to the vertical onto a solution of corn syrup in water. (a) If the beam is refracted to 21.44 to the vertical, what is the index of refraction of the syrup solution? H (b) Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find its wavelength. nm (c) What is its frequency? H: (d) What is its speed in the solution? m/s M H 7. [-/1 Points] DETAILS SERCP7 22.P.010. Light containing wavelengths of 400 nm, 500 nm, and 650 nm is incident from air on a block of crown glass at an angle of 22.0. 1.54 Crown glass 1.52 1.50 Acrylic 1.48 Fused quartz 1.46 400 500 600 700 2, nm Figure 22.14 (a) Are all colors refracted alike, or is one color bent more than the others? O 400 nm light is bent the most O 650 nm light is bent the most O all colors are refracted alike O 500 nm light is bent the most (b) Calculate the angle of refraction in each case to verify your answer. 02 = . (400 nm) 02 = . (500 nm) 02 = . (650 nm)8. [-/1 Points] DETAILS SERCP7 22.P.021. The light beam shown in Figure P22.21 makes an angle of p = 17.50 with the normal line /' in the linseed oil. Determine the angles 0 and @'. (The refractive index for linseed oil is 1.48.) 0 = 0'= Air - Z Linseed oil Water Figure P22.21