Question about light reflection

[kaptinmorgan]

Homework Statement

A diver shines a flashlight upward from beneath the water at a 23.0° angle to the vertical. At what angle does the light leave the water?

Homework Equations

I thought that i could use snell's law here...but i don't know what to do afterwards...

The Attempt at a Solution

angle of incidence is 23... and the index of refraction is 1.333

1.333 (sin 23) = 1 (sin x)

is that about right?

 

[Shooting Star]

angle of incidence is 23... and the index of refraction is 1.333

1.333 (sin 23) = 1 (sin x)

is that about right?

Write very clearly what x represents here. What do you feel that the question wants?

 

[Moetasim]

Your attempt at using Snell's law is correct. However, in order to find the angle at which the light leaves the water, you need to solve for x in the equation you have written. This can be done by taking the inverse sine of both sides of the equation, giving you:

x = sin^-1 (1.333 (sin 23))

Using a calculator, you can find that the angle of refraction is approximately 34.05°. This means that the light will leave the water at an angle of 34.05° from the vertical. This is known as the angle of refraction and is a result of the light bending as it passes from one medium (water) to another (air).