A spotlight on a boat is y = 2.7 m above the water, and the light strikes the water at a point that is x = 8.6 m horizontally displaced from the spotlight (see the drawing). The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom.

find the angle between the beam and the normal. Then, the distance from the boat to the point of incidence (use Tan). Then use snell's law to find the angle of refraction. Then, determine the distance from the incident point by TanAngle=distance/depth

add that distance to 8.6 m.

A drawing will help you greatly.

To find the distance d, we can use the concept of similar triangles. Let's break down the problem using a diagram:

*
/ |
/ |
/ | y = 2.7 m
/ |
/ |
/ |
/ |
/ | d
*-----------------*
/ x = 8.6 m /
/ /
/ /
/ Depth of water = 4.0 m
/ /
/ /
*-------------------*
Bottom of the water

In the diagram, we have a right triangle with sides x, y, and d.

We can set up a proportion using the similar triangles:

x/d = y/(d + 4.0)

Now we can solve for d.

Cross-multiply the equation:

x(d + 4.0) = yd

Expand the equation:

xd + 4.0x = yd

Rearrange the equation to isolate d:

xd - yd = -4.0x

Factor out d:

d(x - y) = -4.0x

Divide both sides by (x - y):

d = -4.0x / (x - y)

Now we can substitute the given values of x, y, and the depth of water, and calculate d:

d = -4.0(8.6) / (8.6 - 2.7)

d = -34.4 / 5.9

d ≈ -5.83 m

Since distance cannot be negative, we disregard the negative sign and take the absolute value:

d ≈ 5.83 m

Therefore, the point where the light strikes the bottom of the water is approximately 5.83 meters away from the boat.