Index of Refraction

  • #1

Homework Statement



2nkrgo.png


Index of refraction of air is being treated as 1, index of refraction for water is 1.33.
W is given as 6.1 cm, the problem is to find H.

Homework Equations



By Snell's Law, n1sin[tex]\Theta[/tex]1=n2sin[tex]\Theta[/tex]2.

We can also use at least one of the trigonometric equations, tan[tex]\Theta[/tex]=opposite/adjacent.

The Attempt at a Solution



Since we know n1 and n2, we can reduce the Snell's Law equation to: sin[tex]\Theta[/tex]1=1.33*sin[tex]\Theta[/tex]2.

We can also get that tan[tex]\Theta[/tex]1=6.1/H and tan[tex]\Theta[/tex]2=3.05/H via basic trig, and these can be combined to get tan[tex]\Theta[/tex]1=2*tan[tex]\Theta[/tex]2.

The problem is that I can't see any way to combine any of those equations that will let me isolate a single variable (H, [tex]\Theta[/tex]1, or [tex]\Theta[/tex]2) so that I can find its value and plug it in to solve for the other variables.
 

Answers and Replies

  • #2
SammyS
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In the figure on the left, the angle from the side of the glass to the ray, is equal to the angle of refraction in the figure on the right. You called this angle, θ1.

Therefore,  [tex]\sin(\theta_1)=\frac{W}{\sqrt{W^2+H^2\,}\,}\ .[/tex]

From the given conditions, it's clear that  [tex]\sin(\theta_2)=\frac{W/2}{\sqrt{(W/2)^2+H^2\,}\,}\ .[/tex]

Since, you know W, n1, and n2, you should be able to write Snell's Law with only one unknown.
 
  • #3
Oh wow, I can't believe I forgot about that! Yeah, that should do it, thanks!
 

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