Index of Refraction!!?

Question

In the setup sketched, you determine the index of refraction of the prism by measuring the two angles θ1 = 61.5 and θ2 = 32.6. Suppose that your error of measurement of each angle is Δθ = 2 degrees. What is the absolute error of the index of refraction?

Answer 1

Hopefully i am understanding the question. Here is my take on it. Absolute error is just the difference between the measured and expected value. So I would just calculate the index of refraction for the angles you have using Snell's law. That number will be your expected value, because that is what you should get if you did nothing wrong. Next I would calculate the index of refraction again using Snell's law but this time ad "2" to both of your angles , because now we are simulating your mistake/error in the lab of being off by 2 degrees. The last step is to simply subtract the two values for the index of refraction and there you go.

Answer 2

Let me guess, this is homework, right? How did I know? Well, there isn't a sketch in your question!

Hint: write out Snell's law. This will give you the index of refraction.

Hint: Find dn (use the quotient rule) with the given values of thew angles and the given uncertainties (remember to convert to radians!).

Answer 3

______________________________
Index of refraction=sin i / sin r

Assuming these are angles of incidence (i = θ1 = 61.5) and angle of refraction( r = θ2 = 32.6),

61.5 degree=1.07337 radian

2 degree =0.0349

Absolute error in sin i= [ |cos i| di / sin i ].

Absolute error in sin 61.5=cos1.07337*.0349 / sin 1.07337= (0.4772*0.0349) /0.8788 = 0.01665/0.8788=0.0189

Absolute error in sin r= [ |cos r| dr / sin r ].

Absolute error in sin 32.6=cos0.569*.0349 / sin0 569= (0.8424*0.0349) /0.5387= =0.05458

Absolute error in measurement of refractive index=Absolute error in 'i' + absolute error. in 'r'=0.0189 +0.05458 =0.0735

Absolute error in measurement of refractive index is 0.0735 0r 7.35%
_______________________________________________
-