Snell's law and inverse function

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SUMMARY

Snell's law describes the relationship between the angles of incidence and refraction when light passes between two media with different refractive indices, expressed as $$\frac{\sin\theta_1}{c_1}=\frac{\sin\theta_2}{c_2}$$. The derived expression for $$\theta_2$$ in terms of $$\theta_1$$ is $$\theta_2 = \sin^{-1}\left(\frac{\sin(\theta_1)}{n_{12}}\right)$$, which is defined only when $$\theta_1 < \arcsin(n_{12})$$. The discussion highlights the importance of understanding the conditions under which total internal reflection occurs, particularly when $$c_2 > c_1$$, leading to the conclusion that $$\theta_1$$ has no upper limit if $$n_{12} > 1$$.

PREREQUISITES
  • Understanding of Snell's Law and its mathematical formulation
  • Knowledge of inverse trigonometric functions, specifically $$\sin^{-1}(x)$$
  • Familiarity with the concept of refractive indices in optics
  • Basic principles of light behavior at media boundaries, including total internal reflection
NEXT STEPS
  • Study the derivation and applications of Snell's Law in various optical scenarios
  • Learn about total internal reflection and its implications in fiber optics
  • Explore the properties of inverse trigonometric functions and their applications in physics
  • Investigate the behavior of light in different media, focusing on refractive index calculations
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Students of physics, particularly those studying optics, educators teaching Snell's Law, and professionals working in fields related to optical engineering and photonics.

Karol
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Homework Statement


Snap1.jpg
Snell's law is:
$$\frac{\sin\theta_1}{c_1}=\frac{\sin\theta_2}{c_2}$$
$$\frac{c_1}{c_2}=n_{12}$$
Express ##\theta_2## as a function of ##\theta_1##
Find the largest value of ##\theta_1## for which the expression for ##\theta_2## that you just found is defined (for larger values of ##\theta_1## than this the incoming light will be reflected).

Homework Equations


Inverse sine: ##y=\sin^{-1}(x)~\rightarrow~\sin(y)=x##

The Attempt at a Solution


$$\sin ( \theta_2 )=\frac{\sin( \theta_1 ) }{n_{12}}~\rightarrow~\sin^{-1}\left( \frac{\sin( \theta_1 )}{n_{12}} \right)=\theta_2$$
##\theta_2## can be ##\frac{\pi}{2}## at the max:
$$\sin^{-1}\left( \frac{\sin( \theta_1 )}{n_{12}} \right)=\frac{\pi}{2}~~\rightarrow~~\sin\left( \frac{\pi}{2} \right)=\frac{\sin(\theta_1)}{n_{12}}$$
$$\Rightarrow~\sin(\theta_1)=n_{12}\sin\left( \frac{\pi}{2} \right)=n_{12}$$
$$\theta_1<\arcsin(n_{12})$$
I didn't use at all the definition of inverse function in the second question, i feel what i have done isn't what it's meant from the chapter of inverse functions
 
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Karol said:
I didn't use at all the definition of inverse function in the second question
Not sure what you mean by that. The answer you gave involved an inverse function, and your final step involved inverting the sine function. Can you be more specific about what it is that you feel you have not used?
By the way, the diagram doesn't match the text. It clearly shows a case with c2>c1, whereas the text assumes the reverse.
 
haruspex said:
the diagram doesn't match the text. It clearly shows a case with c2>c1, whereas the text assumes the reverse.
How does the text show anything about the diagram? only the last formula: ##\theta_1<\arcsin(n_{12})## to my opinion, may show something in that direction.
if ##c_2>c_1~~\rightarrow~\sin(\theta_2)>\sin(\theta_2)~\rightarrow~\theta_2>\theta_1## and the diagram shows the inverse.
The diagram for ##c_2>c_1## must be:
Snap1.jpg

for ##c_2>c_1~\rightarrow~n_{12}<1## and ##\theta_1## has a definite value and that's correct.
 
Karol said:
How does the text show anything about the diagram?
I confused you by incorrectly describing the mismatch.
The diagram shows an example of c1>c2, as you say. But in that case there is no limit (before π/2) on θ1. n12>0, so the arcsin of it is not defined.
Your second diagram, with c2>c1, makes more sense. E.g. total internal reflection can occur from water (low c) to air (high c), but not the other way around.
 
haruspex said:
The diagram shows an example of c1>c2, as you say. But in that case there is no limit (before π/2) on θ1. n12>0, so the arcsin of it is not defined.
Your second diagram, with c2>c1, makes more sense. E.g. total internal reflection can occur from water (low c) to air (high c), but not the other way around.
True, that is also why i thought my answer wasn't right, since there is no limit for ##\theta_1## if ##n_{12}>1##
 

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