A spotlight mounted beside a swimming pool sends a beam of light at an angle int
ID: 1378424 • Letter: A
Question
A spotlight mounted beside a swimming pool sends a beam of light at an angle into the pool. The light hits the water a distance x = 2.4 m from the edge of the pool. The pool is d =2 m deep and the light is mounted h = 1.2 m above the surface of the water. How far from the edge of the pool is the spot of light on the bottom of the pool? (The diagram is not drawn to scale.) Indices of refraction: water: 1.33; air: 1.00029
a-Angle of the beam from the normal in the air?
b-Angle of the beam from the normal in the water ?
c-Distance of spot from edge of pool?
Explanation / Answer
A.
t1 = Arctan (x/h)
Thus,
t1 = 63.435 degrees [ANSWER]
*****************
Note that from Snell's law,
n1sin(t1) = n2sin(t2)
where
n1 = index of refraction of first medium = 1.00029
t1 = angle of incidence = 63.435 degrees
n2 = index of refraction of second medium = 1.33
t2 = angle of refraction =
Thus,
t2 = 42.2756 degrees [ANSWER]
***********************
Distance of spot = x + dtan(t2)
= 4.218 m [ANSWER]
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