(8c34p5) The figure shows a small lightbulb suspended above the surface of the w

Question

(8c34p5) The figure shows a small lightbulb suspended above the surface of the water in a swimming pool, with y1 - 1.6 m and y2 2.2 m. The bottom of the pool is a large mirror. How far below the mirror's surface is the image of the bulb? (cm) (HINT: Costruct a diagram of two rays like that of Fig. 35-3, but take into account the bending of light rays by refraction. Assume that the rays are close to the vertical axis through the bulb, and use the small-angle approximation that sin(e) tan(e) e.) Submit AnswerTries 0/7

Explanation / Answer

Apperent height of bulb above water surface y2' = y2 * n = 2.2*1.33 = 2.926 m

Distance below mirror surface = y1 + y2' = 1.6 +2.926 = 4.526 m answer

For ray diagram, one ray is just vertical extention below the mirror of the line in the image shown.

other ray comes at an angle on water surface, gets refracted and travels in water and then get reflacted at mirror.

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