Apparent depth and snells law

  • Thread starter roam
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  • #1
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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: [tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]

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 [tex]\theta_1 = 90[/tex].

[tex]1= 1.33 sin \theta_2[/tex]

[tex]\theta_2 = 48.75[/tex]

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:

phys.gif


The correct answer is 60 cm.
 

Answers and Replies

  • #2
ehild
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The angle in Snell's law is measured from the normal of the water surface, and it is near zero now.

ehild
 
  • #3
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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?
 
  • #4
ehild
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It is not zero, just small.

ehild
 
  • #5
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Oh, okay. How does this help us to determine the apparent depth?
 
  • #6
ehild
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Look at the figure and find some right triangles.

ehild
 

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