Cylinder i vatten och ljus

(a)An opaque cylindrical tank with an open top has a diameter of 3.15 m and is completely filled with water. When the afternoon sun reaches an angle of 32.5° above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank (in m)? 

Svar: $3.57597448586392$m
Determine how the depth of the tank is related to the angle of refraction for the geometry described in the question. Drawing a diagram may be helpful. Then apply Snell's law to find the angle of refraction and use it to determine the depth of the water in the tank. 

(b)What If? On winter solstice in Missoula, the sun reaches a maximum altitude of 19.7° above the horizon. What would the depth of the tank have to be (in m) for the sun not to illuminate the bottom of the tank on that day? 

Svar: $6.02569903054332$m

Tjena har en fråga har helt fastnat på denna.... Jag använder detta på A Snell's law sin(32.5°) = 1.33 * sin(θ₂) sin(θ₂) = sin(32.5°) / 1.33 = 0.40342917352885 θ₂ = arcsin(0.40342917352885) = 23.7786291969946° djup = (3.15/2) / tan(23.7786291969946°) djup = 3.57597448586392 m och för B använder jag följande Snell's law: sin(19.7°) = 1.33 * sin(θ₂) sin(θ₂) = sin(19.7°) / 1.33 = 0.25408061282075 θ₂ = arcsin(0.25408061282075) = 14.7225669670366° tan(14.7225669670366°) = (3.15/2) / depth djup = (3.15/2) / tan(14.7225669670366°) djup = 6.02569903054332 m