Question

4. y (m) Refer to the diagram (a cutaway side view-not to scale) An enclosed tank, with vertical glass sides and horizontal bottom and top, is nearly full of water. The tank's six panels are all made of the same glass, with uniform thickness d glass of Resting on top of the tank is a solid half-cylinder of glass (with its end facing out of the page), so that its rectangular surface is horizontal and facing upward. The x-y position of the radial center of the half-cylinder's shape is at point P. (with its end facing o air glass Note the coordinate axes given. The origin is defined as point O; the +x-axis is along the tank's inner bottom, and the ty-axis is along the tank's left side, as shown. A laser pointer is aimed (at some non-vertical angle) at point A on the outside bottom of the tank. As a result... Laser light travels through water for a known distance d Later, it travels through air vertically upward to reach the outer surface of the glass half-cylinder (see enlarged detail in lower diagram). At that point, it makes a known angle θ with the glass surface. The ume ΔtG that it takes to travel through the glass half-cylinder is also known. a. Find the coordinates (in cm) of point A glass tank water water laser light x (m) b. Find the coordinates (in cm) of point B, where the laser light reaches the y-axis The data: dlas-1.28 cm dter 46.9 cm glass P (78.6,95.4) cnm 0- 10.70 Δ.-0.425 ns E laser E light n , = 1.52 water glass n1.00 c-3.00 x 108 m/s ı (enlarged detail) ir

Answer 1

Distance xy = c/n_glass times Delta t_G xy = 3 times 10^8/1,52 times 0.425 times 10^- 9 = 8.4 cm from Snell�s law at x 1 times sin (90 - 10.7) = 1.52 sin theta_1 theta_1 = 40.3 degree theta_2 = 180 - 90 - theta - theta_1 theta_2 = 39 degree sine law R/sin (90 + theta_2) = py/sin theta_1 = xy/sin theta R/sin (129 degree) = py/sin (40.3) = 8.4/sin (10.7) R = 35.16 cm py = 29.26 cm Snell�s law at y n_glass sin (theta_2) = sin (theta_3) trheta_3 = 73.05 degree theta_3 = 73.05 degree f = R sin theta = 35.16 times cos (10.7) = 34.55 cm \r\n Snell�s law m_water sin theta_4 = 1 times sin (90 degree) sin theta_4 = (1/1.52) theta_4 = 41.14 degree n_water sin (90 - theta_4) = n_glass sin theta_6 theta_6 = 41.2 degree