Solve Index of Refraction Homework Statement

In summary, the problem involves finding the height (H) using the index of refraction for air and water (1 and 1.33 respectively), and a given width (W) of 6.1 cm. The solution involves using Snell's Law and trigonometric equations to find the value of H. The key equation is sinθ1 = W/√(W^2+H^2), and with the given values, Snell's Law can be written with only one unknown.
  • #1
ObliviousSage
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Homework Statement



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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 into solve for the other variables.
 
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  • #2
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!
 

1. What is the index of refraction?

The index of refraction is a measure of how much a material slows down the speed of light passing through it. It is represented by the symbol "n" and is specific to each material.

2. How is the index of refraction calculated?

The index of refraction is calculated by dividing the speed of light in a vacuum by the speed of light in the material. This can also be written as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material.

3. What does a high index of refraction indicate?

A high index of refraction indicates that the material slows down the speed of light more than a material with a lower index of refraction. This is typically seen in dense materials such as glass or water.

4. How does the index of refraction affect the path of light?

The index of refraction determines the angle at which light bends when it enters a new material. This is known as refraction and is what causes objects to appear distorted when viewed through a lens or a glass of water.

5. What factors can affect the index of refraction?

The index of refraction can be affected by several factors, including the density of the material, the wavelength of the light passing through it, and the temperature of the material. Additionally, impurities or defects in the material can also affect the index of refraction.

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