Snell's law question with a twist....

[L-x]

Homework Statement

https://gyazo.com/f8cf156e7bd2f2511e3fa859e3732fe6

Homework Equations

Snell's law.

The Attempt at a Solution

I'm mostly confused about the relevance of the second medium (of index n2) given that the distance between the ring and the boundary is "small". Having attempted to find an epxression for the apparent depth as a function of d which ignored the second boundary except that the observed knew the ring was next to it (and therefore that the apparent depth was L/tan(Θ_0) - h, I ended up with the apparent depth being equal to
d/n1 cos(Θ_1) / cos(Θ_0), the rate of change of which is only 0 when cos (Θ_1) is 0, i.e. the ring is on the surface of the water. This is clearly wrong, but I'm a loss at how to include the information about a second medium. Ray diagrams assuming the ring is some distance δ<<L into the second medium have left me with more unknowns and further from a solution...

Edit: it looks like I at least didn't make a mistake with the formula for apparent depth ignoring the second medium, as some googling provided me with a formula for apparent depth in terms of depth and theta_1 which is a trivial re-arrangement of the equation I mentioned above.

 

[kuruman]

The apparent depth formula that you quoted is not independent of the actual depth. You are looking for a solution where the apparent depth becomes independent (at some point) of how far down the ring sinks. How can that happen? Hint: What happens to a ray that is incident at the critical angle for total internal reflection between media 2 and 1? Draw a ray diagram of the path it would follow until it emerges from the surface of medium 1.

 

[L-x]

Thanks! I can see now that as the ring falls deeper, the angle light from the ring makes with the boundary between the two media approaches 90, and therefore the angle that the light ray makes with the other side of the boundary approaches the critical angle.