What is the angle of refraction if a ray of light passes from water into air at an angle of 30 degrees?

Apr 24, 2014

One problem solving strategy that I find very useful in answering a question like this is GRASS.

<b>G - what am I Given? </b> Assign each value to a number.
${n}_{1}$ = the index of refraction in the first substance, which is water (1.33)
${\Theta}_{1}$ = the angle of incidence (${30}^{o}$).
${n}_{2}$ = the index of refraction in the second substance, which is air (1.0003)

<b>R - what am i Required to find? </b> This is my unknown (${\Theta}_{2}$).

<b>A - what formula do I need to use? Do I need to manipulate the equation to isolate my unknown value?</b>
To answer a question like this, use Snell's Law (${n}_{1} \sin {\Theta}_{1} = {n}_{2} \sin {\Theta}_{2}$). The unknown in this problem is ${\Theta}_{2}$, which should be isolated.

<b>S - Solve.</b>
$\sin {\Theta}_{2} = \frac{{n}_{1} \cdot \sin {\Theta}_{1}}{n} _ 2$
$\sin {\Theta}_{2} = \frac{1.33 \cdot \sin {30}^{o}}{1.0003}$
$\sin {\Theta}_{2} = 0.6648$
${\Theta}_{2} = {\sin}^{-} 1 0.6648$
${\Theta}_{2} = {41.66}^{o}$

<b>S - Significant digits. </b>
Since ${30}^{o}$ only contains 2 significant digits, the answer needs to be rounded to 2 significant digits, which means the angle of refraction in air is ${42}^{o}$.