# Apparent depth

1. Aug 16, 2010

### roam

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

A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly above the fish? (For water, n = 1.33.)

2. Relevant equations

Snell's law: $$n_1 sin \theta_1 = n_2 sin \theta_2$$

3. The attempt at a solution

So far I have found the angle of refraction in the water using Snell's law. Since they are asking for "a position almost directly above the fish", I took $$\theta_1 = 90$$.

$$1= 1.33 sin \theta_2$$

$$\theta_2 = 48.75$$

But I what else can I do?? I'm really stuck! I need to find the difference between he apparent and the real depth x, 80-x=apparent length. But how? Here's a diagram which shows a similar situation:

The correct answer is 60 cm.

2. Aug 16, 2010

### ehild

The angle in Snell's law is measured from the normal of the water surface, and it is near zero now.

ehild

3. Aug 16, 2010

### roam

If this angle is zero, then what do I need to do to find the apparent depth?

4. Aug 16, 2010

### ehild

It is not zero, just small.

ehild

5. Aug 16, 2010

### roam

Oh, okay. How does this help us to determine the apparent depth?

6. Aug 16, 2010

### ehild

Look at the figure and find some right triangles.

ehild