Question

Light is incident on the surface of water in an aquarium with an angle of 36 degrees. It travels through the water, reflects off the mirror on the bottom of the aquarium and emerges back into the air. If the aquarium is 1 m deep and light travels at a speed 2.22 x10^8 m/s in the aquarium, determine the distance between the light's entry and exit points. (Hint: the aquarium water contains many dissolved substances and will have a slightly different index of refraction than regular water)

Answer 1

where is your solution?

Answer 2

I'm not sure how to solve it since i've never done a question like it. I think that I would need to use the formula n1sin(theta)1=n2sin(theta)2 and sub in the index of refractions for air and water and sub in the angle 36 degrees into sin(theta) for water. I also think that I would have to find the wavelength for the distance between the points. I don't know if I am supposed to do that though nor what do with the speed and depth

Answer 3

With the speed given you can determine the index of refraction of the water in the aquarium.

\(\eta=\dfrac{c}{v_{medium}}\)

Answer 4

From here you can see that angle a=90\(^o\)-b