# A scuba diver uses his waterproof flashlight to shine a beam of light so that it strikes the surface of the water at an angle of incidence θi. Use Snell’s law to find the angle of incidence that...

- A scuba diver uses his waterproof flashlight to shine a beam of light so that it strikes the surface of the water at an angle of incidence θi. Use Snell’s law to find the angle of incidence that would give an angle of refraction for the refracted ray to be directed right along the surface, and show that θi is the same as the critical angle for total internal reflection.

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The Snell's Law states

`sin(theta_i)/sin(theta_r) = n_2/n_1` ,

where `theta_i` is the angle the incident ray of line makes with the line perpendicular to the surface, `theta_r` is the angle the refracted ray of light makes with the line perpendicular to the surface, `n_r` is the index of refraction in the refractive medium (in this case,air) and `n_i` is the index of refraction of incident medium (in this case, water).

For air, the index of refraction is `n_r = 1` .

Since the refracted ray is to be directed along the surface, it will make the angle of 90 degrees with the line perpendicular to the surface. Thus, `sin(theta_r) = sin(90) = 1`

Plugging these values in the Snell's Law, calculate the sine of the incident angle:

`sin(theta_i) = 1* 1/1.33 =0.75 `

This gives the incident angle approximately 49 degrees.

To understand why this is the same angle as the critical angle for total internal reflection, rewrite the Snell's Law as

`sin(theta_r) = n_i/n_r *sin(theta_i)`

It can be seen from here that the greater the `sin(theta_i)` , the greater the `sin(theta_i)` . Since sine is the increasing function, increasing the incident angle beyond the value which results in `sin(theta_r)` = 1 will result in `sin(theta_r)` greater than 1 which is impossible. The light will not refract but reflect back into the incident medium. So 49 degrees is the critical angle for total internal reflection.

(This only happens when `n_i/n_r > 1` , that is, the incident medium has a higher refractive index. Otherwise `sin(theta_r)` would remain less than 1.)