Question 1163329
I agree with ikleyn and CPhill, but I would like to draw and overexplain as usual.
{{{drawing(350,300,-3,4,-1,5,
blue(line(-3,2,4,2)),line(-3,0,4,0),
line(-2.6,3.5,0,2), green(line(0,2,3.464,0)),
line(0,2,1.716,0),rectangle(0,0,0.2,0.2),
red(line(0,0,0,5)),arc(0,2,3,3,-150,-90),
arc(0,2,3,3,49.37,90),locate(-0.7,3.2,theta),
locate(0.25,1,theta),locate(-0.54,3.1,1),
locate(0.45,0.9,2),circle(-2.6,3.5,0.1),circle(1.716,0,0.1),
green(circle(3.464,0,0.1)),arc(0,2,3.4,3.4,30,90),
locate(1.4,1,theta),locate(1.6,0.9,1),
locate(-2.8,4.9,eye),arrow(-2.6,4.6,-2.6,3.6),
locate(-0.9,1.2,2ft),arrow(-1,1.3,-1,0),arrow(-1,0.7,-1,2),
arrow(1.9,-0.2,0,-0.2),arrow(1.5,-0.2,3.464,-0.2),
arrow(1.3,-0.6,0,-0.6),arrow(0.5,-0.6,1.716,-0.6),
locate(0.3,-0.6,1.716ft),locate(2,-0.2,3.464ft),
arrow(1.716,-0.8,1.716,-0.1),locate(1.5,-0.7,goggles),
green(arrow(3.464,2.9,3.464,0.1)),locate(3.1,3.1,green(ghost)),
locate(2.8,3.4,green(goggles))
)}}}
 
{{{drawing(20,30,0,0.4,-0.6,0,
locate(0.39,0,theta)
)}}}{{{1}}}{{{"="}}}{{{60^o}}} (given)
{{{drawing(20,30,0,0.4,-0.6,0,
locate(0.39,0,theta)
)}}}{{{2}}} value calculated based on Snell's Law: {{{"sin("}}}{{{theta}}}{{{"1)"}}}{{{"="}}}{{{n*"sin("}}}{{{theta}}}{{{"2)"}}}
{{{sin(60^o)}}}{{{"="}}}{{{1.33*"sin("}}}{{{theta}}}{{{"2)"}}}-->{{{sqrt(3)/2}}}{{{"="}}}{{{1.33*"sin("}}}{{{theta}}}{{{"2)"}}}-->{{{sqrt(3)/2/1.33}}}{{{"="}}}{{{"sin("}}}{{{theta}}}{{{"2)"}}}
From that we can find {{{theta}}}{{{"2)"}}}{{{"="}}}{{{highlight(40.63^o)}}} (rounded)
We can also find {{{"cos("}}}{{{theta}}}{{{"2)"}}} ,
{{{"tan("}}}{{{theta}}}{{{"2)"}}}, and the distance
{{{2ft*"tan("}}}{{{theta}}}{{{"2)"=highlight(1.716ft)}}} (rounded)
Independently of refraction, we can find the distance
{{{2ft*"tan("}}}{{{theta}}}{{{"1)"}}}{{{"="}}}{{{2ft*tan(60^o)=2ft*sqrt(3)=highlight(3.464ft)}}} (rounded)