Asked by physics
The critical angle of refraction for calcite is 68.4 degrees when it forms aboundary with water. Use this information to determine the speed of light in calcite.
refaraction index 'n'
n=cosec 68.4degrees
n=1.07
is this done correctly?
refaraction index 'n'
n=cosec 68.4degrees
n=1.07
is this done correctly?
Answers
Answered by
drwls
They asked you for a speed of light, not an "n". If 1.07 is supposed to be the index of refraction for calcite, it is incorrect.
When a beam of light leaving calcite forms a critical angle at a water boundary, the angle of refraction is A2 = 90 degrees. Using Snell's law with N2 = 1.33 for water,
N1 sin A1 = N2 sin A2 = N2
N1 = 1.33/sinA1 = 1.43
speed of light in calcite
= (3*10^8 m/s)/1.43 = ?
When a beam of light leaving calcite forms a critical angle at a water boundary, the angle of refraction is A2 = 90 degrees. Using Snell's law with N2 = 1.33 for water,
N1 sin A1 = N2 sin A2 = N2
N1 = 1.33/sinA1 = 1.43
speed of light in calcite
= (3*10^8 m/s)/1.43 = ?
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