The figure shows a small lightbulb suspended at distance d 1 = 260 cm above the

Question

The figure shows a small lightbulb suspended at distance d1 = 260 cm above the surface of the water in a swimming pool where the water depth is d2 = 270 cm. The bottom of the pool is a large mirror. How far below the mirror surface is the image of the bulb? (Hint: Construct a diagram of two rays like that of figure (b) below, but take into account the bending of light rays by refraction at the water surface. Assume that the indices of refraction of the air and the water are 1.00 and 1.33 respectively, and the rays are close to a vertical axis through the bulb, and use the small-angle approximation in which sinO = tanO=O.)

Please show work and type out final answer

The figure shows a small lightbulb suspended at distance d1 = 260 cm above the surface of the water in a swimming pool where the water depth is d2 = 270 cm. The bottom of the pool is a large mirror. How far below the mirror surface is the image of the bulb? (Hint: Construct a diagram of two rays like that of figure (b) below, but take into account the bending of light rays by refraction at the water surface. Assume that the indices of refraction of the air and the water are 1.00 and 1.33 respectively, and the rays are close to a vertical axis through the bulb, and use the small-angle approximation in which sinO = tanO=O.)

Explanation / Answer

apparent deapth = real deapth/refractive index of water

let Distance below mirror surface that light bulb is d

d = 260 + 270/1.33 = 463 cms (answer)

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