Question

A submarine is 282 m horizontally from the shore of a freshwater lake and 110 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 209 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level. (a) Draw a diagram of the situation, identifying the two triangles that are important in finding the solution. (b) Find the angle of incidence of the beam striking the water-air interface. ° (c) Find the angle of refraction. ° (d) What angle does the refracted beam make with the horizontal? ° (e) Find the height of the target above sea level.

Please include a drawing of the diagram.

Answer 1

(a) From draw the diagram D air 0 0 water ө A B (b) From the figure, AB = distance from the point of incidence of light to the sunmarine = 282 - 209 = 73 m BO=depth of the submarine =110 m The angle of incidence at the surface of water is, AB O= tan-1 BO 73 -1 = tan 110 = 33.57° (c) Using Snell's law n, sin = Nair sino Ø=sin-1 n, sin e nar 1.33 sin (33.57°) = sin = 47.34° (d) The angle made by the refracted beam with the horizontal is, O=90-47.34 42.66°

(e) The height of the target above sea level is, h=(OC)tan(90 - $) = (209) tan ) tan(47.34°) = 226.80