Index of refraction in an air bubble

AI Thread Summary
The discussion focuses on calculating the apparent depth of an air bubble within a plastic ball using the index of refraction. The bubble is located 2.90 cm from the surface of a 9.00 cm-diameter ball. The equation used is n_1/s + n_2/s' = n_2-n_1/R, leading to the conclusion that the calculated value for s' is 0.024 m, indicating the bubble appears virtually beneath the surface. The final clarification confirms that the value is correct, and since s' is virtual, it should be considered negative. The solution effectively demonstrates the application of optical principles in determining perceived depth.
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[SOLVED] Index of refraction

Homework Statement


An air bubble inside an 9.00 cm-diameter plastic ball is 2.90 cm from the surface.

As you look at the ball with the bubble turned toward you, how far beneath the surface does the bubble appear to be?


Homework Equations


n_1/s + n_2/s' = n_2-n_1/R


The Attempt at a Solution


1.59/.0290 + 1/s' = 1-1.5/.045
s'=1.52

I'm not sure why that isn't correct.

...

1.59/.0290+1/s' = 1.59-1/.045
s'=.024m <- correct
 
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Here s' is virtual. So it should be -ve.
Your value is correct.
 
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