Can you please help with these questions #1-6. The questions are in the screenshots down below.

#1-3 are in the screenshots below:

1. [-/1 Points] DETAILS MY NOTES SERCP7 22.AE.02. ASK YOUR TEACHER ' PRACTICE ANOTHE Example 22.2 Angle of Refraction for Glass Goal Apply Snell's law to a slab of glass. Incident N | Norma Problem A light ray of wavelength 589 nm (produced by a sodium Tay lamp) traveling through air is incident on smooth, flat slab of crown glass at an angle of &, = 24 to the normal, as sketched in Figure 4 22.11. Find the angle of refraction, &,. 0 et I I I I I | Strategy Substitute quantities into Snell's law and solve for the N | I unknown angle of refraction, 8. Air : = . va 1 X Refracted ray Figure 22.11 Refraction of light by glass. Solve Snell's law (Eq. 22.8) for sin 8, sinfs = sinf); (1) ng 1.00 From Table 22.1, find n; = 1.00 for air and n, = 1.52 dinde (1 _9) (sin24.0) = 0.268 J for crown glass. Substitute these values into (1) and 02 take the inverse sine of both sides. 8, = sin"1(0.268) = :] o 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 24.0 to the normal, determine the angle of refraction. The ray bends ( away from the normal, as expected. 2. [-/1 Points] DETAILS MY NOTES SERCP7 22.AE.04. ASK YOUR TEACHER | PRACTICE ANO1 [~ \\J 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 ny passes through a thick transparent slab with parallel faces and index of refraction n, (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 8; and 65. Figure 22.12 When light passes through a flat slab of material, the emerging beam is parallel to the incident beam, and therefore 6, = 6;. Apply Snell's law to the upper surface. sinfly n_lsinf)l (1) ny Apply Snell's law to the lower surface. sinf "_2511192 (2) n Substitute Equation 1 into Equation 2. . n ny . 4 4 singy = (lsma,) ny \ 2 Take the inverse sine of both sides, noting that the 63 select 6 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. I [ | Exercise 22.4 Hints: Getting Started | I'm Stuck [' ' Suppose the ray, in air with n = 1.00, enters a slab with n = 2.48 at a 33.3 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. [-/1 Points] DETAILS MY NOTES SERCP7 22.AE.06. ASK YOUR TEACHER | PRACTICE ANO1 a A\\ Example 22.6 A View from the Fish's Eye Goal Apply the concept of total internal reflection. Problem (a) Find the critical angle for a water-air boundary if the index of refraction of water is 1.33. (b) Use the result of part (a) to predict what a fish will see (Fig 22.28) if it looks up toward the water surface at angles of 40.6, 48.8, and 58.7. Strategy After finding 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. (a) Find the critical angle for a water-air boundary. Substitute into Equation 22.9 to find the critical angle. (b) Predict what the fish will see if it looks up toward the surface at angles of 41, 49, and 59. 40,6 Direction 48.8 Direction 58.7 Direction Figure 22.28 A fish looks upward toward the water's surface. 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 reflection of some object on the bottom. II A beam of light from underwater will be refracted 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 air-water boundary without changing direction. The fish 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. f | Exercise 22.6 Hints: Getting Started | I'm Stuck | 1 critical angle for total internal reflection for light traveling in the oil layer and encountering the oil-water boundary? ~ 6. = Suppose a layer of oil with n = 1.45 coats the surface of the water. What is the | ( 4. [-/1 Points] DETAILS MY NOTES SERCP7 22.P.009. ASK YOUR TEACHER PRACTICE ANOTHER A laser beam is incident at an angle of 37.0 to the vertical onto a solution of corn syrup in water. (a) If the beam is refracted to 25.84 to the vertical, what is the index of refraction of the syrup solution? J (b) Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find its wavelength. | nm (c) What is its frequency? Hz (d) What is its speed in the solution? | m/s Submit Answer 5. [-/1 Points] DETAILS MY NOTES SERCP7 22.P.010. ASK YOUR TEACHER PRACTICE ANOTHER 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 28.0. i et | 1,5.|t' | . ' "Crown g':l.rii 50 - 1.50 Acrvlic _ e 1.48 |- 1.46 i . I O 400 500 600 700 A, nm Figure 22.14 (a) Are all colors refracted alike, or is one color bent more than the others? () 400 nm light is bent the most (O 500 nm light is bent the most (O 650 nm light is bent the most () all colors are refracted alike (b) Calculate the angle of refraction in each case to verify your answer. 8, = (400 nm) 6; = (500 nm) 8, = (650 nm) 6. [-/1 Points] DETAILS MY NOTES SERCP7 22.P.021. ASK YOUR TEACHER PRACTICE ANOTHER The light beam shown in Figure P22.21 makes an angle of p = 23.0 with the normal line /' in the linseed oil. Determine the angles 0 and 0'. (The refractive index for linseed oil is 1.48.) 0 = Air N Linseed oil - Water Figure P22.21