College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- As we saw in the video in step 1, total internal reflection can occur for cases where a light ray attempts to travel from a high index to low index medium. However, it can only occur for particular ranges of incident angles. The lowest possible incident angle for which total internal reflection occurs is called the critical angle. With = 1.50 and n₂ = 1.00, use the simulation to find the critical angle (in degrees). 1₂ 7₂. 1' Play around with different values of ₁, ₂, and 8₁. Try to find the critical angle for different values of n₁ and Note that total internal reflection is only possible in cases where n₁ > ₂.arrow_forwardCalculate the critical angle for the light traveling from glass (n glass= 1.54) to air (n=1.00). Express your answer in degrees. Normal format with 3 SF. airarrow_forward6. A sheet of glass has an index of refraction nalass = 1.45. Assume that the index of refraction for air is na= 1.00, Light is incident on the sheet of glass at an angle of incidence of 60°, as shown in the figure below. a. Calculate and label the size of each angle (in degrees) on the figure, including angles of incidence and refraction at each of the two parallel surfaces shown. 60 degrees Air na = 1.00 Glass " = 1.45 Air ng = 1.00 b. Calculate the velocity of the light as it travels in the glass.arrow_forward
- A ray of light travels through air until it strikes the interface between the air and another medium. The incident ray makes an angle of 0, = 32.0° with the normal, as shown in the figure below. Upon passage into the second medium, the ray is refracted, emerging from the interface at an angle 0, with respect to the normal. Air Second medium (a) Suppose that the second medium is polystyrene. What is the angle of refraction, 8, (in degrees)? (Enter your answer to at least one decimal place.) (b) Suppose that the second medium is sodium chloride. What is the angle of refraction, 8,, in this case (in degrees)? (Enter your answer to at least one decimal place.) (c) Finally, suppose that the second medium is glycerine. What is the angle of refraction, 0,, in this case (in degrees)? (Enter your answer to at least one decimal place.)arrow_forwardThe critical angle for total internal reflection at a liquid-air interface is 45.0°. Part A If a ray of light traveling in the liquid has an angle of incidence at the interface of 34.0 °, what angle does the refracted ray in the air make with the normal? Express your answer in degrees. ΑΠΑΣΦ 0 = Submit Part B Request Answer 0 = ? If a ray of light traveling in air has an angle of incidence at the interface of 34.0 °, what angle does the refracted ray in the liquid make with the normal? Express your answer in degrees. —| ΑΣΦ 2 ? Oarrow_forwardE. Light is incident from air (n = 1.0003) on the rectangular piece of glass shown below. 60° 30° 55° 35° 1. Determine the index of refraction of the glass Determine the speed of light in the glass shownarrow_forward
- Problem 34.51 It's nighttime, and you've dropped your goggles into a 3.0-m-deep swimming pool. If you hold a laser pointer 1.2 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 1.9 m from the edge. Part A How far are the goggles from the edge of the pool? Express your answer with the appropriate units. G ΜΑ ? d = Value Units Submit Request Answer Return to Assignment Provide Feedback a 4 বর DII DD F5 F6 F7 F8 F9 F10 80 F3 F4 $ 54 #3 W E % 85 96 & 27 R T Y U D F G 8 9 ) 0 0 Parrow_forward9arrow_forwardA beam of light is incident from air on the surface of a liquid. If the angle of incidence is 36.2° and the angle of refraction is 25.0°, find the critical angle for the liquid when surrounded by air. Step 1 When light goes from one material into another having a higher index of refraction, the light bends toward the normal line as shown in part (a) of the diagram. 02 air liquid (a) (b) We are given that when = 36.2°, the angle of refraction in the liquid is = 25.0°. Thus, from Snell's law, the index of refraction of the liquid is nliquid = nair sin sin e = (1.00) sin sin 25.0° 0 (0.423)arrow_forward
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