Solve Water Refraction: Length of Stick's Shadow

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SUMMARY

The problem involves calculating the length of the shadow cast by a 1.93-meter tall stick submerged in water, with sunlight entering at a 45-degree angle. Using Snell's Law, where the refractive indices are na=1.00 for air and nb=1.33 for water, the correct angle of refraction is determined to be approximately 32.8 degrees. The shadow length is then calculated using the tangent function, resulting in a final shadow length of 1.45 meters. The initial calculations were incorrect due to misapplication of angles and trigonometric functions.

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[SOLVED] Water Refraction

Homework Statement



An opaque stick which is d = 1.93 meters tall stands vertically upright on the bottom of a pool of clear water. Parallel rays of sunlight making an angle of 45.0 degrees with the water?s surface illuminate the pool as shown. What is the length of the shadow the stick casts on the bottom of the pool?

Homework Equations



[tex]n_{a}[/tex]=1.00
[tex]n_{b}[/tex]=1.33
stick_in_water.gif


The Attempt at a Solution



So I just took

[tex]arctan(\frac{1.33}{1.00})=53.06[/tex]

And I need the angle that will form a triangle with the rod, so

[tex]90-53.06=36.94[/tex]

Then, since the shadow forms the base of a triangle, I solved for it:

[tex]tan(36.94)=\frac{opp}{adj}=\frac{opp}{1.93m}[/tex]

Adn I got that the shadow equals 1.45 m, which was incorrect. What am I doing wrong?
 
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ttiger2k7 said:
So I just took

[tex]arctan(\frac{1.33}{1.00})=53.06[/tex]
Why?

Hint: Use Snell's law to find the angle the light makes upon refraction.
 

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