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

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SUMMARY

The critical angle of refraction for calcite is 68.4 degrees when in contact with water, which is essential for calculating the speed of light in calcite. The correct approach involves using Snell's Law to determine the index of refraction, which is calculated as 1.44. Subsequently, the speed of light in calcite is derived using the formula v = c/n, resulting in a speed of approximately 2.08 x 10^8 m/s. Understanding the relationship between critical angle and refractive index is crucial for solving this type of refraction problem.

PREREQUISITES
  • Understanding of Snell's Law
  • Knowledge of critical angle and its significance
  • Familiarity with the index of refraction
  • Basic proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study Snell's Law in detail to understand its applications in optics
  • Research the concept of critical angle and its implications in different media
  • Learn about the index of refraction for various materials
  • Explore advanced optics problems involving refraction and reflection
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Students studying physics, particularly those focusing on optics, as well as educators and anyone interested in the principles of light behavior in different media.

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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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