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?

1 answer

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 = ?