Answer to Question #92281 in Optics for jane montgomery

Question

A chunk of zircon is sitting in an open container of benzene. A 589nm wavelength of light has gone through the zircon and is emerging into the benzene. It strikes the boundary at an angle of 10.0° from the normal.
How would it be different if the light entered the benzene at a greater angle?
How would it be different if the benzene at an angle greater than the critical angle?

Answer 1

Determine the critical angle of incidence for zircon and benzene.

$$\frac{\ sinθ }{\ sin90°} = \frac{n_{2}}{n_{1}} (1)$$

where n₂=n_benzene=1.5, n₁=n_zircon=1.94

Using (1) we get:

θ=51°

How would it be different if the light entered the benzene at a greater angle?

Answer:

If the angle of incidence is less than 51° the light will move in the same way as at the angle of incidence 10°. At the angle of incidence 51°, the refraction angle will be equal to 90°, that is, the light will move along the boundary between the two media.

How would it be different if the benzene at an angle greater than the critical angle?

Answer:

The light does not go in benzene, but returns in zircon.