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In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point d = 2.9 m from his foot at the edge of the pool Where does the spot of light hit the bottom of the h = 2.2 m deep pool? Measure from the bottom of the wall beneath his foot. Express your answer to two significant figures and include the appropriate units.

Sagot :

The spot of light hit the bottom of the h = 2.2 m deep pool will hit the bottom at 4.98 m away.

Here ,

height h = 2.2 m

h1 = 1.3 m

d = 2.9 m

let the distance of spot where light hit bottom from side is l

angle of incidence , i  = arctan(2.9/1.3)

i = 65.9 degree

Using Snell's law

1 * sin(i) = 1.33 * sin(r)

1 * sin(65.9) = 1.33 * sin(r)

r = 43.4 degree

Now ,  tan(r) = (l - 2.9)/h

tan(43.4 degree) = (l - 2.9)/2.2

solving for x

l = 4.98 m

the spot Light will hit the bottom at 4.98 m away

Learn more about Light here:

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