A researcher stands at the end of a lakeside pier,4.0 m above the water surface

Question

A researcher stands at the end of a lakeside pier,4.0 m above the water surface and 2d= 12 m away from a buoy floating on the lake in (Figure 1) . A rope hangs from the bottom of the buoy down into the water, with a light attached to its end. When the researcher looks at the light from a position at which his line of sight intersects the water surface exactly halfway between the pier and the buoy, the light appears to be d below the water surface.

How far under the surface is the light?

Express your answer with the appropriate units.

Explanation / Answer

from the figure

angle of refraction, theta_r = tan^-1(6/4)

= 56.3 degrees

Let theta_i is the angle of incidense.

n1 = 1.33( for water)

n2 = 1 (for air)

Apply, Snell's law

sin(theta_i)/sin(theta_r) = n2/n1

sin(theta_i) = sin(theta_r)*n2/n1

= sin(56.3)*1/1.33

= 0.6256

theta_i = sin^-1(0.6256)

= 38.73 degrees

Let h is the actula depth

tan(38.72) = d/h

h = d/tan(38.72)

= 6/tan(38.72)

= 7.48 m <<<<<<<--------------Answer

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