Question

A ray of light passes from air into cubic zirconia at an angle of 56.0° to the normal. The angle of

refraction is 22.00°.
What is the index of refraction of cubic zirconia?

Answer 1

η = 2.2

Step-by-step explanation:

The index of refraction is given by the following formula:

`\eta = (Sin\ \theta_i)/(Sin\ \theta r)`

where,

η = index of refraction of cubic zirconia = ?

`\theta_i` = angle of incidence = 56°

`\theta_r` = angle of refraction = 22°

Therefore,

`\eta = (Sin\ 56^o)/(Sin\ 22^o)\\\\\eta = (0.829)/(0.375)`

η = 2.2