# Search light !

1. Apr 2, 2009

### itryphysics

1. The problem statement, all variables and given/known data

A search light is being used at the edge of a tank to find a lost key at the bottom of the tank. The tank is filled to the top with a fluid for which the index of refraction is 1.68. The key is 8.176 m below the surface and 6.84 m from the edge of the tank. At what angle to the normal will the searchlight have to point in order to locate the key if the light enters the water 1 m from the edge of the tank?

2. Relevant equations

snell's law
n1 sin theta1 = n1 sin theta2

3. The attempt at a solution

inverse sin of 6.84/8.176

inverse sign of 7.84/8.176

and then i added those two together..

Im totally lost and dont understand what to do . Any help will be much appreciated =]

2. Apr 2, 2009

### LowlyPion

Draw a picture first off.

The θ2 is given by tan-1 of (6.84 - 1)/8.176

The rest should be easy peasy.

3. Apr 2, 2009

### itryphysics

so everything i did was ok except for I use inverse of tan and not sin to find the angles?

4. Apr 2, 2009

### LowlyPion

Not exactly. You failed to account for the light entering the water at 1m.

In the drawing then your net x displacement over the y depth describes θ with respect to the vertical.

To determine the θ1 of course you use sin-1

5. Apr 2, 2009

### itryphysics

im lost. sorry . so what is wrong with the way i was calculating theta 1?

6. Apr 2, 2009

### LowlyPion

You use Snell's law. So knowing θ2 from the tan-1 and ...

n1*sinθ1 = n2*sinθ 2

1*sinθ1 = 1.68*sinθ2

To determine θ1 you take the sin-1.

Isn't that what you were doing before?