Question

4. a. A ray of light in water (n = 1.33) encounters an interface with a piece of glass (n = 1.65). If the "grazing angle” (angle to the interface) is 30° in the water, calculate the angle of refraction of the ray in the glass. Sketch the ray in both media, correctly showing its "bend.” (Optional: Sketch the reflected ray, too.) 30° nwater = 1.33 n glass = 1.65 angle of refraction in glass: A. 37.4 B. 39.5° C. 41.6 D. 44.3° E. 46.1° b. If the ray of light starts inside the glass and emerges into the water at an angle 30° to the interface, find the angle of incidence of the ray in the glass. (Hint: This should require no new calculation after part (a)!) Sketch the ray in both media, correctly showing its "bend.” (Optional: Sketch the reflected ray, too.) nwater = 1.33 30° nglass = 1.65 angle of incidence in glass: A. 37.4° B. 39.5° C. 41.6 D. 44.3° E. 46.1°

Answer 1

1.33 1.65 Apply Snell's lane 1.33 Sin (60) 1.65 sinor Snor - 1.33 sin 160) - 0.699 Ora sin (0.698) = 44.266° ~ 4.3 degrees option (D) conect 1-33 Apply Snell's lan 1.65 sino = 1.33 sin (60) Sina, - 1.33 sin (G) 1.65 O = sin (0.698) - 44.266 degrees. 8, 8 44.30 = 44.3 degrees option (D) conect