An opaque cylindrical tank with an open top has a diameter of 3.00m and is completely filled with water. When the setting sun reaches an angle of
above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank?
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Let's analyze this optics problem. We have a cylindrical tank with diameter 3.00 meters, completely filled with water. When the sun is at 23.0 degrees above the horizon, sunlight stops illuminating the bottom. This happens when the sun's rays entering at the top edge reach exactly the opposite bottom edge.
The geometry forms a right triangle. The horizontal side is the tank diameter of 3.00 meters, the vertical side is the unknown depth h, and the angle with the horizontal is 23.0 degrees. Using trigonometry, we can write: tangent of 23.0 degrees equals h divided by 3.00 meters.
Now let's solve for the depth step by step. We start with tangent of 23.0 degrees equals h over 3.00. Rearranging gives us h equals 3.00 times tangent of 23.0 degrees. The tangent of 23.0 degrees is approximately 0.42447. Multiplying 3.00 by 0.42447 gives us 1.27341 meters. Rounding to three significant figures, the tank depth is 1.27 meters.
Let's verify our solution. With a tank diameter of 3.00 meters and sun angle of 23.0 degrees, we calculated a depth of 1.27 meters. To check: tangent of 23.0 degrees should equal 1.27 divided by 3.00, which gives 0.423. This closely matches the actual tangent value of 0.424, confirming our answer. Therefore, the tank depth is 1.27 meters.
To summarize this optics and trigonometry problem: We identified that when sunlight stops illuminating the tank bottom, it creates a critical ray forming a right triangle. The horizontal side is the tank diameter, the vertical side is the depth, and we know the sun's angle. Using the tangent function, we derived the formula h equals d times tangent of theta. Substituting our values gives us the final answer: the tank depth is 1.27 meters.