Answer:
therefore critical angle c= 69.79°
Explanation:
Canola oil is less dense than water, so it floats over water.
Given [tex]n_{canola}= 1.47[/tex]
which is higher than that of water
refractive index of water [tex]n_{water}=1.33[/tex]
to calculate critical angle of light going from the oil into water
we know that
[tex]sinc= \frac{n_{water}}{n_{canola}}[/tex]
now putting values we get
[tex]sinc= \frac{1.33}{1.47}[/tex]
c= [tex]sin^{-1}(\frac{1.33}{1.47} )[/tex]
c=69.79°
therefore critical angle c= 69.79°
