# Optics: Distance of Key in swimming pool

[SOLVED] Optics: Distance of Key in swimming pool

1. Homework Statement

After a long day of driving you take a late-night swim in a motel swimming pool. When you go to your room, you realize that you have lost your room key in the pool. You borrow a powerful flashlight and walk around the pool, shining the light into it. The light shines on the key, which is lying on the bottom of the pool, when the flashlight is held 1.2 m above the water surface and is directed at the surface a horizontal distance of 1.5 m from the edge

If the water here is 4.0 m deep, how far is the key from the edge of the pool?

2. Homework Equations
I'm not sure, but I used tan(x)= opposite/adjacent and Snell's Law, nsin(x)=nsin(x)

3. The Attempt at a Solution

The light forms two triangles, so I figured I needed to find the angle of the flashlight to the water first. That means tan(x)=1.2/1.5, and arctan(1.2/1.5) is 38.65 degrees.

From here, I used Snell's Law to figure out the angle in the water, and since the problem didn't state an index of refraction for water, I assumed the standard of 1.33 and used n=1 for air. This means sin(38.65)=1.33sin(x), or arcsin(sin(38.65)/1.33) which equals 28.01 degrees, which I rounded to 28 degrees. This meant the third angle of the underwater triangle is 62 degrees.

That means that to find the length of the bottom side of the triangle b, I needed to use tan(62)= 4/b, or b= 4/tan(62), which I solved to equal approximately 2.13 meters.

So, I figured that the total distance to the key is the horizontal distance of the flashlight plus the underwater distance. So I have 2.13m+1.5m= 3.63 meters, which is wrong according to the computer.

Any help would be greatly appreciated, please!

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2. Homework Equations
I'm not sure, but I used tan(x)= opposite/adjacent and Snell's Law, nsin(x)=nsin(x)

3. The Attempt at a Solution

The light forms two triangles, so I figured I needed to find the angle of the flashlight to the water first. That means tan(x)=1.2/1.5, and arctan(1.2/1.5) is 38.65 degrees.

From here, I used Snell's Law to figure out the angle in the water, and since the problem didn't state an index of refraction for water, I assumed the standard of 1.33 and used n=1 for air. This means sin(38.65)=1.33sin(x),
The angle of incidence is the angle the light ray makes with the normal to the surface. So it should be (90-38.65)° which you should be using in Snell’s law.

Ah.... stupid mistakes like that always throw me off...

Thanks a bunch!