Question

A 1.8 cm thick layer of oil (n = 1.46) floats on top of 33 cm deep water (n = 1.33) in a rectangular bucket. When a laser beam is held at « = 37 degrees with respect to the surface of the water, its beam enters at the edge of the bucket and strikes the base of the bucket at its center. oil w a. Re-create the figure on your page and draw the path of the light as it passes through the oil and water. Clearly label all angles and distances that you use in your work. b. At what angle with respect to the surface of the water does the light travel in the oil? c. At what angle with respect to the surface of the water does the light travel in the water? d. How wide is the bucket? Submit this answer below.

Answer 1

A Oil (n=1:46) J.scm c B 09 Waler (n = 1:33) 33cm M N W W given Q=37° Incidence angle of the laser beam at the air- oil inter face ()= 0 = 9029 90°- 37° Angle of refraction of the beam from the air- oil interface is a From Bigure. / BAC = LACE is a. Aljernate angles So incident angle of the beam of oil wafer Vinterface Angle of refraction of the beam from oil-wafer interface is on Since the lasere hits the base of the bucket af its centre. Hence wa (width of the bucket is w DN =

Refractive index of oil ie refractive index of oil with respect to air is noil Noil nair 1.46 Refractive index of water is nwater - nwater = 1:33 nain nualen Nwaten naire X Nain noil 1:33 1:46 noil medium of reefractive index (1.1) n O We know, Snell's law of reefraction Sino na n Interface Sinop two medium medium of rei na Applying Snell's law at aire-oil interface noil Sino Sing Maire Sin 53 1.46 SMO s Sin + Sin53 1.6 33.16216. 0 33:16 nwater Noil Snelles law at oil-wafer interface Sino Sino Sin 33.16° 1:33 Singa 1:46 2.46 Sin 33:16° 1:33 -/ Sin

= 36.9016 - ř 36.9 from sigure, AB = 18 cm BD = CM = 33 cm For A ABC BC tano, AB BC = AB tano, For ACMN MN tano cm MN = CM fano DN = DM + MN Be + MN AB tano, + CM tano, 1:8 (tan 33-160) + 33/tan 36.99) cm cm 25.95 3/9 - 25.953 cm Since DN = 4/2 W = 2 DN = (2x 26963) om 51.906 cm The light travel in the oil with respect to the Sunface of the water at an angle P = 90°-Q = 90°- 33.16° = 56.84°

The light travel in the water with respect to the surface of the water at an angle Q2 = 9000 - 90°- 36.99 53.1° The width of the bucket is W = 51.906 cm