Finding speed of light through medium, when given critical angle of refraction

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Homework Help Overview

The problem involves determining the speed of light in calcite using the critical angle of refraction when it interfaces with water. The critical angle provided is 68.4 degrees, which is relevant to the calculation of the index of refraction and subsequently the speed of light in the medium.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the critical angle and the index of refraction, with some attempting to derive the refractive index from the critical angle using trigonometric relationships. Questions arise regarding the role of water in the calculations and the correct application of Snell's Law.

Discussion Status

The discussion is active, with participants providing different approaches to calculate the refractive index and the speed of light in calcite. Some guidance has been offered regarding the need to correctly interpret the critical angle and its implications for the calculations, but there is no explicit consensus on the correct method yet.

Contextual Notes

Participants note the importance of understanding the definitions and relationships involved in the problem, particularly the distinction between angles and indices of refraction. There is also mention of the potential irrelevance of the angle of incidence in the context of the current calculations.

seanmcgowan
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Homework Statement


The critical angle of refraction for calcite is 68.4 (degrees) when it forms a boundary with water. Use this information to determine the speed of light in calcite.



Homework Equations


n=c/v
n= Index of refraction
c= Speed of light in a vacuum
v= Speed of light in medium

The Attempt at a Solution



I am looking for v so i rearranged the equation to look like this: v=c/n

v= (3x10^8)/68.4= 4385964.91228

Is this right? Did the water have something to do with the equation? If not then could you please help, I have no idea how to work this problem otherwise.
 
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First, and perhaps most significantly, 68.4 degrees is an angle... not an index of refraction (which you have plugged it in as). You also should look up what "critical angle" means... since you will need to use this angle as one of the quantities in your calculation... a calculation that will probably first yield index (which you can then use in the equation you've shown). Hint: also look up information about "Snell's Law"... this is a refraction problem after all, and critical angle has to do with a critical point in the math of Snell's law.
 
So here's what i did:

using: sin(critical angle)= index of refraction for second medium/ Index of refraction for first medium...

sin(68.4)= 1.333/x
x=1.33/sin(68.4)= 1.44

then using the the n=c/v equation:

1.44= 3x10^8/x
x=3x10^8/1.44= 208333333.33

so would the answer be 2.08x10^8? also, i didn't find any use for snell's law. it would have given me the angle of incedence, which i believe is irrelevant, but i may be wrong.
 
using the previous eq. u must find refractive index which is sin(angle of incidence)/sin(angle of refraction) then use the refractive index to find the speed of light in another medium by doing this:
v=c/n
where c=3x10^8
and n is the refractive index you found earlier.
Hope this helps
 

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