Question

A giant ocean tank at an aquarium has acrylic plastic walls 18 cm thick. The index of refraction of acrylic plastic is 1.49. A fish is 220 cm from the inside wall. To a viewer on the outside, how far does the fish appear to be from the outside wall? Hint: The of the first refraction is the object for the second refraction.

Accepted Answer

Identify the key concept: This problem involves refraction and the apparent position of an object when viewed through a medium with a different index of refraction. We'll use the formula for refraction at a flat surface: $$ s' = \frac{s}{n} $$, where $$ s  is $$the object distance, $$ s'  is $$the apparent distance, and $$ n  is $$the index of refraction of the medium. Step 1: Calculate the apparent position of the fish as seen from the inside wall of the acrylic plastic. The fish is located $$ s = 220 \; 	ext{cm} $$ from the inside wall, and the index of refraction of acrylic plastic is $$ n = 1.49 . $$Using the formula $$ s' = \frac{s}{n} $$, substitute the values to find the apparent position of the fish inside the acrylic plastic. Step 2: Recognize that the apparent position of the fish inside the acrylic plastic (calculated in Step 1) becomes the object distance for the second refraction, as the light exits the acrylic plastic into air. The index of refraction of air is approximately $$ n = 1.00 . $$Step 3: For the second refraction, the object distance is the thickness of the acrylic wall (18 cm) minus the apparent position of the fish inside the acrylic (from Step 1). Use the same refraction formula $$ s' = \frac{s}{n} $$, where $$ s  is $$the new object distance and $$ n  is $$the index of refraction of acrylic plastic, to calculate the apparent position of the fish as seen by the viewer outside the tank. Step 4: Add the thickness of the acrylic wall (18 cm) to the apparent position of the fish (from Step 3) to determine the total apparent distance of the fish from the outside wall as seen by the viewer.

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