Question #92281

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?

Expert's answer

Determine the critical angle of incidence for zircon and benzene.


 sinθ sin90°=n2n1(1)\frac{\ sinθ }{\ sin90°} = \frac{n_{2}}{n_{1}} (1)

where n2=nbenzene=1.5, n1=nzircon=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.


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