Answer:
therefore critical angle c= 69.79°
Step-by-step explanation:
Canola oil is less dense than water, so it floats over water.
Given
![n_(canola)= 1.47](https://img.qammunity.org/2020/formulas/physics/college/7ufrgb55pb5h0xz038h1rp8jz4wv0vwc9n.png)
which is higher than that of water
refractive index of water
![n_(water)=1.33](https://img.qammunity.org/2020/formulas/physics/college/fneq0a1z7oxwabvgc59i3c1mlf94fwhhcz.png)
to calculate critical angle of light going from the oil into water
we know that
![sinc= (n_(water))/(n_(canola))](https://img.qammunity.org/2020/formulas/physics/college/y0ifnc96q9jgbeinmbjmlen3ww3ivuu1ri.png)
now putting values we get
![sinc= (1.33)/(1.47)](https://img.qammunity.org/2020/formulas/physics/college/65y7r9nlbj8n7a77dwo7sco5lohx5j7ixc.png)
c=
![sin^(-1)((1.33)/(1.47) )](https://img.qammunity.org/2020/formulas/physics/college/yi0b1kzade1f2mi2uttzcmhshjvxjrkcel.png)
c=69.79°
therefore critical angle c= 69.79°