Calculating Apparent Depth Using Snell's Law

[roam]

Homework Statement

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.)

Homework Equations

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

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.

 

[ehild]

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

ehild

 

[roam]

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

ehild

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

 

[ehild]

It is not zero, just small.

ehild

 

[roam]

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

 

[ehild]

Look at the figure and find some right triangles.

ehild