QUESTION 5 For problem 33.107 calculate the index of refraction of the glass if the refracted angle is 38.8 degrees. 5 sig. figs.

A ray of light travelling through air encounters a 1.7-cm-thick sheet of glass at a 45 ∘ angle of incidence. Assume n = 1.5. How far does the light ray travel inside the glass before emerging on the far side?Express your answer in centimeters.

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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.

(a) An opaque cylindrical tank with an open top has a diameter of 2.60 m and is completely filled with water. When the afternoon sun reaches an angle of 31.5° above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank (in m)? 4 Determine how the depth of the tank is related to the angle of refraction for the geometry described in the question. Drawing a diagram may be helpful. Then apply Snell's law to find the angle of refraction and use it to determine the depth of the water in the tank. m (b) What If? On winter solstice in Seattle, the sun reaches a maximum altitude of 19° above the horizon. What would the depth of the tank have to be (in m) for the sun not to illuminate the bottom of the tank on that day? m

In the above figure, theta (the angle the incident ray makes with respect to the vertical) is 26.0 degrees. What is d (the distance between the ray emerging from the bottom of the glass and where the ray would have been if it had continued straight on with no glass to refract it)?

A beam of light shines on an equilateral glass prism at an angle of 41° to one face. 1) What is the angle at which the light emerges from the opposite face? nglass = 1.57 (Express to 3 sig fig)

Compute the minimum deviation angle of a refracting prism of index of refraction 1.7 if the apex of the prism is 70 degrees.

An object is placed 9.50 cm in front of the cornea. (The cornea is thin ans has approximately parallel sides so that the reflection that occurs as light travels from air to cornea to aqueous humor is essentially the same as though the aqueous humor were directly in contact with the air. The aqueous humor has index of refraction n = 1.34 and the radius of curvature of cornea is 7.8 mm.) (a) What is the image distance for the image formed by the cornea alone? (b) The image formed by the cornea serves as an object for the lens. Treat the lens as a thin lens 4 mm behind the cornea. Find the optical power of the lens necessary to form an image on the retina, 23 mm from the center of the lens.