Doppler Shift when Light Travels Through Two Different Mediums When considering the Doppler shift, the 'canonical equation' is 
$$f=\frac{c+vr}{c+vs}f_0$$
However, this equation seems to run into trouble in the following situation: A light source inside water is moving at a speed $v$ towards a receiver outside the water. Can we modify the above equation to deal with a transition between two media? Or is there a completely different formula that applies here?
 A: If we assume that an observer in the water and one in the air are not moving with respect to each other or the water boundary, then they will both measure the same frequency.   When the wave hits the water boundary, the waves can't "pile up" there.  So the same number that arrive each second must leave each second.
I believe therefore that you could do this as two calculations.
One in water to calculate the observed frequency at the water/air boundary, and a second calculation for the detector to the boundary (assuming there is relative motion between the two).
For light, we don't consider there to be a distinct propagation medium that the receiver and sender have motion relative to.  (Yes, you have a speed in the water, but the speed of light is not fixed relative to the water).  So that simplifies the formula:
$$ f = f_0 \frac{1}{1 + \frac{v}{c}}$$

Example: emitter is moving at $80 m/s$ to the right, and the viewer is moving at $400m/s$ to the right.  The boundary is stationary.  The emitted frequency is $6.0\times 10^{14} Hz$.  We'll take the index of refraction of water at that frequency to be $1.33$.
$$f = 6.0\times 10^{14}Hz \frac{1}{1 + \frac{-80\frac{m}{s}}{\frac{c}{1.33}}}$$
$$f = 6.0\times 10^{14}Hz \frac{1}{1 + (-3.55 \times 10^{-7})}$$
$$f = 6.0000021 \times 10^{14}Hz$$
Since the object is approaching the boundary, the light is blue shifted.  Now the shift for the observer:
$$ f = 6.0000021 \times 10^{14}Hz \frac{1}{1 + \frac{400\frac{m}{s}}{c}}$$
$$ f = 6.0000021 \times 10^{14}Hz \frac{1}{1 + (1.33 \times 10^{-6})}$$
$$ f = 5.9999941 \times 10^{14}Hz$$
The observer is moving away and the light is redshifted (by a greater amount).
