Question

With what angle should a ray strike the surface of water to make an angle of refraction 20 degree in water? The refractive index of water is 1.33.

[Answer: 27.057]​

Answer 1

Approximately
`27^(\circ)`, assuming that the light entered the water from the air (
`n_(i) \approx 1`.)

Step-by-step explanation:

In this question, the following is given:

• Angle of refraction:
  `\theta_(r) = 20^(\circ)`, and
• Refractive index of the new medium (water):
  `n_(r) = 1.33`.

Assuming that the the ray of light entered water from the air:

• Refractive index of the original medium (air):
  `n_(i) \approx 1`.

Apply Snell's Law to find the angle of incidence,
`\theta_(i)`.

By Snell's Law, in a refraction:

`n_(i)\, \sin(\theta_(i)) = n_(r)\, \sin(\theta_(r))`. where:

• 
  `n_(i)` is the refractive index of the original medium,
• 
  `\theta_(i)` is the angle of incidence (before entering the new medium,)
• 
  `n_(r)` is the refractive index of the new medium, and
• 
  `\theta_(r)` is the angle of refraction (after entering the new medium.)

Rearrange this equation to find the angle of incidence
`\theta_(i)`:

`\displaystyle \sin(\theta_(i)) = \left((n_(r))/(n_(i))\right)\, \sin(\theta_(r))`.

`\begin{aligned} \theta_(i) &= \arcsin\left(\left((n_(r))/(n_(i))\right)\, \sin(\theta_(r))\right) \\ &\approx \arcsin\left(\left((1.33)/(1)\right)\, \sin(20^(\circ))\right) \\ &\approx 27^(\circ)\end{aligned}`.

(Note that the angle is measured in degrees.)

In other words, the beam of light should enter the water at approximately
`27^(\circ)` from the normal.