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?
Yes, your calculation is correct. The refractive index of a material, denoted by "n," is defined as the ratio of the speed of light in vacuum to the speed of light in the material. In this case, you calculated the refractive index of calcite by taking the reciprocal of the sine of the critical angle of refraction, which is 68.4 degrees.
So, the refractive index of calcite is approximately 1.07. Keep in mind that this value may be rounded off, and the speed of light in calcite can be determined by dividing the speed of light in vacuum (usually denoted as "c") by the refractive index.
Yes, your calculation is correct. The refractive index of a medium, denoted by 'n', is equal to the reciprocal of the sine of the critical angle for refraction. In this case, the critical angle of refraction for calcite is given as 68.4 degrees.
To determine the refractive index, you can use the formula:
n = 1/sin(critical angle)
Substituting the given angle value:
n = 1/sin(68.4) ≈ 1.07
Thus, the value you have calculated for the refractive index of calcite is correct.
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 = ?