A rock is dropped in a large tank of benzne (n = 1.5) and sinks to the bottom of

Question

A rock is dropped in a large tank of benzne (n = 1.5) and sinks to the bottom of the tank. If the apparent depth of the rock as viewed directly from above is 35 cm, what is the actual depth of the rock? Hint: See above probem.

Above problem:

A more general analysis of refracting surfaces would include the case of refraction of light at a spherical interface such as a fishbowl. Similar to the thin-lens equation, the expression relating the object distance, image distance, indices of refraction, and the radius of curvature of the surface is given by n1/so + n2/si = (n2-n1)/R where n1 and n2 are indices of refraction to the left and right of the refracting surace, respectively. The radius of the curvature of the bowl is R = - 17 cm and the fish is at the center of the bowl, 8 cm from the curved surface and 6 cm down from the top. We can neglect the glass bowl and only consider the refraction due to the water. Determine the image distance of the fish as viewed from a) the side and b) the top. You will need to determine the radius of curvature of a flat surface to answer part b.

Explanation / Answer

Apparent depth=h*(nair/nbenzene)

35=h*1/1.5

h=52.5cm

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