Question

What is the critical angle between polycarbonate (n1 = 1.58) and quartz (n2 = 1.41)?

24 degrees
46 degrees
63 degrees
75 degrees

Answer 1

Approximately
`63.2^(\circ)`.

Step-by-step explanation:

The refractive index
`n` gives the ratio between the speed of light in vacuum and the speed of light in the given material. A larger refractive index means slower speed of light in the medium. In this question, the speed of light is slower in polycarbonate than in quartz.

When light travels from a material of low light speed to a material of higher light speed, the angle of refraction
`\theta_(r)` will exceed the angle of incidence
`\theta_(i)` (both with respect to the normal.)

The critical angle at this interface is the
`\theta_(i)` for which
`\theta_(r) = 90^(\circ)`. Note that since
`\theta_(i)\!` cannot exceed
`90^(\circ)`,
`\theta_(i) < \theta{r}` and light must be entering a faster medium from a slower medium. In this question, light would be entering quartz (
`n = 1.41`, faster) from polycarbonate (
`n = 1.58`, slower.)

By Snell's Law:

`n_(i)\, \sin(\theta_(i)) = n_(r)\, \sin(\theta_(r))`,

Where:

• 
  `n_(i)` is the refractive index of the medium from which light enters,
• 
  `n_(r)` is the refractive index of the medium light enters into,
• 
  `\theta_(i)` is the angle of incidence at which light enters the interface (with respect to the normal), and
• 
  `\theta_(r)` is the angle of refraction at which light leaves the interface (also with respect to the normal.)

Set
`n_(i) = 1.58` (polycarbonate),
`n_(r) = 1.41` (quartz), and
`\theta_(r) = 90^(\circ)`. Rearrange the equation and solve for
`\theta_(i)`:

`\begin{aligned}\sin(\theta_(i)) &= (n_(r)\, \sin(\theta_(r)))/(n_(i)) \end{aligned}`.

`\begin{aligned}\theta_(i) &= \arcsin\left ((n_(r)\, \sin(\theta_(r)))/(n_(i)) \right) \\ &= \arcsin \left((1.41\, \sin(90^(\circ)))/(1.58)\right) \\ &\approx 63.2^(\circ)\end{aligned}`.