A spotlight on a boat is y = 2.6 m above the water, and the light strikes the water at a point that is x = 8.6 m horizontally displaced from the spotlight. The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom.
m
Find the angle of incidence (arctan8.6/2.6)
then use snells law to find the angle of refraction.
I am not certain if d is the distance from the boat...assuming so...
d= 8.6+4ctn(anglerefraction)
checkthat.
To determine the distance d, you can use the concept of similar triangles.
The triangle formed by the spotlight, the point where the light strikes the water, and the point where the light strikes the bottom of the water is similar to the triangle formed by the spotlight, the point where the light strikes the water, and the bottom of the water.
Let's call the distance from the spotlight to the point where the light strikes the water "a" and the distance from the spotlight to the bottom of the water "b". Now, using the similar triangles, we can set up the following equation:
a / (a + b) = y / (y + 4.0)
Plugging in the given values, we have:
(a) / (a + b) = 2.6 / 6.6
To solve for "a", cross multiply:
2.6(a + b) = 6.6a
2.6a + 2.6b = 6.6a
2.6b = 4.0a
b = (4.0 / 2.6) * a
Now, we know that the point where the light strikes the water is horizontally displaced by 8.6 m from the spotlight. Therefore, we can set up another equation:
a + b = 8.6
Substituting the value of "b" from the previous equation:
a + (4.0 / 2.6) * a = 8.6
(1 + 1.54) * a = 8.6
2.54 * a = 8.6
a = 8.6 / 2.54
Now, we can plug this value of "a" back into the equation for "b":
b = (4.0 / 2.6) * a
Calculate the value of "b" using the value of "a" you just obtained. This will give us the distance from the spotlight to the bottom of the water.