53. A light ray in water passes into air. The angle ofincidence in the water is

Question

53. A light ray in water passes into air. The angle ofincidence in the water is 40 degrees. What is the angle ofrefraction (in degrees) in air if the index of refaction of wateris 1.33.    a. 29    b. 40    c. 59    d. .59 54. Diamond has an index of refraction of 2.42. The criticalangle for diamond in air:    a. does not exist since the index is greaterthan two    b. would only exist if the diamond werecoated with a transparent material    c. is about 24 degrees    d. is about .4 degrees 55. A pool is filled to a depth of 2.0 meters. How deep doesthe pool appear to be from above the water?    a. -2.0 m    b. -1.5 m    c. 1.5 m    d. 1.33 m 53. A light ray in water passes into air. The angle ofincidence in the water is 40 degrees. What is the angle ofrefraction (in degrees) in air if the index of refaction of wateris 1.33.    a. 29    b. 40    c. 59    d. .59 54. Diamond has an index of refraction of 2.42. The criticalangle for diamond in air:    a. does not exist since the index is greaterthan two    b. would only exist if the diamond werecoated with a transparent material    c. is about 24 degrees    d. is about .4 degrees 55. A pool is filled to a depth of 2.0 meters. How deep doesthe pool appear to be from above the water?    a. -2.0 m    b. -1.5 m    c. 1.5 m    d. 1.33 m

Explanation / Answer

angle of incidence i = 40 degrees from snell's law sin i / sin r = 1 / 1.33 from this angle of refraction r = 58.74degrees (b). Index of refraction of diamond n = 2.42 So, critical angle for daimond in air   C = sin-1 ( 1/ n )                                                           = 24.4 degrees (c). apprent depth = real depth / refractive index ofwater                             2 m / 1.3333                            = 1.5 m

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