Change of angle if refractive index changes for 10^-6

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The discussion centers on the calculation of the change in angle when the refractive index changes, specifically using the formula (delta (ϑ)) = -(minus delta n) /(n*sqrt((n^2-1))). Participants confirm the correctness of the solution while seeking clarification on the steps involved in taking differentials. The method involves applying differentiation to functions like sin(θ) and 1/n to find their respective changes. One participant expresses initial confusion but ultimately resolves their query with assistance. The thread concludes with a successful resolution of the problem.
AncientOne99
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Homework Statement
The following expression sin ϑ = 1 / n applies to the angle of total reflection at the transition of the beam from glass to air. For how much
changes the angle if the refractive index changes by 10^−6?
For n take 1.5. Express the result only with n.
Relevant Equations
sin(ϑ) = 1/n(water)
Solution:
(delta (ϑ)) =
-(minus delta n) /(n*sqrt((n^2-1)))
 
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If you are asking whether your answer is correct, it is.
 
Yes i know the solution, but i don't know the steps.
 
Do you know how to take differentials? For example
##d(x^2)=\frac{d}{dx}(x^2)dx=2xdx.##

Apply this idea to find ##d(\sin\theta)## and ##d(\frac{1}{n})## and set them equal.
 
kuruman said:
Do you know how to take differentials? For example
##d(x^2)=\frac{d}{dx}(x^2)dx=2xdx.##

Apply this idea to find ##d(\sin\theta)## and ##d(\frac{1}{n})## and set them equal.
Yes, i know how to take the diferentials, but there is no equation to diferentiate or i don't know it.
 
I solved it, thanks for your help !
 
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