Question

A ray of monochromatic light ( f = 5.09 × 10^14 Hz) passes from air into Lucite at an angle
of incidence of 30.˚.

Using a protractor and straightedge, on the diagram, draw the refracted ray in the
Lucite.

Answer 1

     Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:

$\frac{sen\O_{2}}{sen\O_{1}} = \frac{n_{2}}{n_{1}} \\ sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}}$
 
     Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:

$sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}} \\ sen\O_{2}= \frac{1.5* \frac{1}{2} }{1} \\ sen\O_{2}=0.75$
  
     Using the arcsin properties, we get:

$sen\O_{2}=0.75 \\ arcsin(0.75)=\O_{2} \\ \boxed {\O_{2}=48.59^o}$

Obs: Approximate results, and the drawing is attached

If you notice any mistake in my english, let me know, because i am not native.