Ascuba 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.

Snell's law: n1Sinα=n2Sin β where α=Incidence angle, β=angle of refraction, n1 and n2 are the indices of refraction for water and air respectively.
Therefore,
Sinα=n2/n1 Sinβ For refracted ray to be along the surface of water, β=90° and thus Sinβ = 1

Sinα=n2/n1= 1/1.33 = 0.7519 => α=sin^-1 (0.7519) = 48.75°
When light moves from a medium of higher index of refraction (such as water) to medium of lesser index of refraction (such as air), the refracted ray is bend such that α is bigger than β. This  is internal refraction. At some value of α, β approaches 90°. This incidence angle is called critical incidence angle. Therefore, the current scenario is shows critical angle of incidence.

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