4. A light ray is incident on a glassinterface (n = 1.5). It refracts throughthe medium and then exits back out intothe air as shown in the picture.Determine the final refracted angle 0, asthe light ray exits the glass. 0₁ = 18° and=34°.

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Answered: 4. A light ray is incident on a glass…

College Physics

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Chapter1: Units, Trigonometry. And Vectors

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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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4. A ray of light passes through a prism, making the minimum angle of deviation. The rays enter and emerge at 7° from the normal to the surface. What is the index of refraction of the material, if the prism angle is 50°?
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11:19 Optics.pdf -> 2. Find angle ø. (Hint: Use Geometry skills.) 79.4712° Air, n, = 1 3. Find the second angle of 30 incidence. 10.5288° Horiz. ray, parallel to 4. Find the second angle of refraction, 0, using Snell's Law 15.9° base Glass, n = 1.5 Refraction Problem #2 Goal: Find the distance the light ray displaced due to the thick window and how much time it spends in the glass. Some hints given. 20° 0, 1. Find 0. 20° H,0 2. To show incoming & outgoing rays are parallel, find 0. 20° n, = 1.3 3. Find d. 0.504 m glass 10m 4. Find the time the light spends in the glass. n = 1.5 5.2 · 10-8 s Extra practice: Find 0 if bottom medium is replaced with air. H,0 26.4° Refraction Problem #3 Goal: Find the exit angle relative to the horizontal. 0 = 19.8° 36 air glass The triangle is isosceles. Incident ray is horizontal, parallel to the base.
4. In a church choir loft, two parallel walls are 5.30 m apart. The "singers stand against the north wall. The organist faces the south wall, sitting 0.800 m away from it. So that she can see the choir, a flat mirror 0.600 m wide is mounted on the south wall, straight in front of the organist. What width of the north wall can she see? Hint: Draw a top-view diagram to justify your answer.
A1. As shown in Figure A1, a light ray is incident normal to one face of a 30°-60°-90° prism (n = 1.50) that is immersed in water (n = 1.33). Determine the exit angle 03. (a) Determine the exit angle 03. (b) Determine the critical angle for the prism-water system. n2 60.0° n1 Figure Al 021 30.0% D 103
2. A ray of light in glass (n₁=1.5) enters another material (n2) at a 70° angle; and emerges into the new material at a 45° angle. Find the index n2. n₁=1.5 70° N2 45°
1. A ray of light is passing through air (n=1). It strikes a level surface of water (n=1.33) at an angle of 40° with respect to the surface of the water. The water is 10 cm thick. The light ray refracts and travels through the water. There is a level layer of clear polystyrene (n = 1.6) that the water is resting upon. The light ray hits the polystyrene level and travels through it, finally reaching a sheet of glass (n=1.45). Given that the speed of light in a vacuum is 3 x 108 m/s, calculate the speed of light in the water and the time it takes for the light to traverse the water layer. (actual question i need answered) *Using the arrangement in Problem 1, calculate the critical angle for total internal reflection at the polystyrene-glass interface
30.0° 1. A light ray passes into a rectangular piece of glass (n = 1.50) as shown. How much does the path of the light shift (d) once it emerges? 2.00 cm

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4. A light ray is incident on a glass interface (n = 1.5). It refracts through the medium and then exits back out into the air as shown in the picture. Determine the final refracted angle 0, as the light ray exits the glass. 0₁ = 18° and =34°.