Can you please help me with these 6 questions. I am very confused and I do not understand these questions. The screenshots of these questions are down below.

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1. [-/1 Points] DETAILS MY NOTES SERCP7 22.AE.02. ASK YOUR TEACHER PRACTICE ANOTHER [~ R T, Goal Apply Snell's law to a slab of glass. Incident N | Norma Ay Problem A light ray of wavelength 589 nm (produced by a sodium . lamp) traveling through air is incident on smooth, flat slab of crown glass at an angle of 8; = 24 to the normal, as sketched in Figure 22.11. Find the angle of refraction, &,. Strategy Substitute quantities into Snell's law and solve for the unknown angle of refraction, &5. K 1\\ 1 03 '. Relracted ray Figure 22.11 Refraction of light by glass. Solve Snell's law (Eq. 22.8) for sin 0, sinfly = SiIll (1) na for crown glass. Substitute these values into (1) and 5H2 take the inverse sine of both sides. 6, = sin"1(0.268) = ( ]o L From Table 22.1, find n; = 1.00 for air and n, = 1.52 dindy = (I.UU) (sin24.0) = 0.268 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 : angle of 24.0 to the normal, determine the angle of refraction. ( || If the light ray moves from inside the glass toward the glass-air interface at an ' [ | | g The ray bends away from the normal, as expected. | ' I { 1 2. [-/1 Points] DETAILS MY NOTES SERCP7 22.AE.04. ASK YOUR TEACHER PRACTICE ANOTHER B 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 &5. Figure 22,12 When light passes through a flat slab of material, the emerging beam is parallel to the incident beam, and therefore ; = 6;. Apply Snell's law to the upper surface. sinfs ging! 1e (1) N2 Apply Snell's law to the lower surface. sinf Esiuu {2) Ty Substitute Equation 1 into Equation 2. : ny fny . 4 1 sinfly = =2 (lsuw,) ny ng Take the inverse sine of both sides, noting that the ) 5, 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. | Exercise 22.4 Hints: Getting Started | I'm Stuck 4 ! l 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? 8= J