Finding a Lost Key with a Searchlight: Calculate the Angle

[itryphysics]

Homework Statement

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?

Homework Equations

snell's law
n1 sin theta1 = n1 sin theta2

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 don't understand what to do . Any help will be much appreciated =]

 

[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.

 

[itryphysics]

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

 

[LowlyPion]

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

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

 

[itryphysics]

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

 

[LowlyPion]

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

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?