Question

If the light strikes the plastic (from the water) at an angle θw, at what angle θa does it emerge from the plastic (into the air)? Express your answer in terms of nw, np, na, and θw. Remember that the inverse sine of a number x should be entered as asin(x) in the answer box.

Answer 1

Answer:

$\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})$

Explanation:

According to Snell's law the angle of incidence and angle of refraction are related by:

$\frac{sin(\theta _w)}{sin(\theta _a)}=\frac{n_a}{n_w}\\\\\therefore sin(\theta _a)=\frac{sin(\theta _w)\times n_w}{n_a}\\\\\theta _a=sin^{-1}(\frac{sin(\theta _w)\times n_w}{n_a})\\\\\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})$

Answer 2

Answer:

θa = asin(nwsin(θw)/na)

Explanation:

We are going to look at the two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface). In order to ascertain their relationship, draw a picture with the path the light follows in the plastic and the normals to both surfaces. Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normals are parallel lines. An important theorem from geometry will give you the relationship between the angles.

Using Snell's Law,

θa = asin(nwsin(θw)/na)