Question

Canola oil is less dense than water, so it floats on water, but its index of refraction is 1.47, higher than that of water. When you are adding oil and water to a bottle to make salad dressing, you notice a silvery reflection of light from the boundary between the oil and water. What is the critical angle for light going from the oil into the water?

Answer 1

Answer:

therefore critical angle c= 69.79°

Explanation:

Canola oil is less dense than water, so it floats over water.

Given $n_{canola}= 1.47$

which is higher than that of water

refractive index of water $n_{water}=1.33$

to calculate critical angle of light going from the oil into water

we know that

$sinc= \frac{n_{water}}{n_{canola}}$

now putting values we get

$sinc= \frac{1.33}{1.47}$

c= $sin^{-1}(\frac{1.33}{1.47} )$

c=69.79°

therefore critical angle c= 69.79°