Understanding Snell's Law: Common Mistakes and Troubleshooting Tips

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Homework Help Overview

The discussion revolves around Snell's Law and the conditions for total internal reflection in optics. Participants are examining the relationships between the angles of incidence and refraction, as well as the refractive indices involved in the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply Snell's Law and are questioning their understanding of the equations related to total internal reflection. There is a focus on the inconsistency arising from the equations used, particularly regarding the angle of incidence and the critical angle.

Discussion Status

The discussion is ongoing, with participants expressing confusion about their calculations and the relationships between the angles and refractive indices. Some guidance has been offered regarding the conditions for total internal reflection, but no consensus has been reached on the correct approach.

Contextual Notes

Participants note that they have two unknowns to solve for and are grappling with the implications of total internal reflection on their equations. There is mention of a figure attached to the problem statement, which may contain relevant information that is not fully articulated in the text.

jegues
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Homework Statement



See figure attached for problem statement.

Homework Equations





The Attempt at a Solution



I'm getting a disagreement in my equations so I must be doing something wrong.

We have two unknowns,

n_{1},\theta_{2}

My first equation,

n_{1}sin\theta_{1}=n_{2}sin\theta_{2}

my second equation (This is where I think I am misunderstanding something)

Since it is experiencing total internal reflection,

\theta_{2} = arcsin\frac{n_{1}}{n_{2}}

Or in other words,

sin\theta_{2} = \frac{n_{1}}{n_{2}}

If I plug this into my first equation I get an inconsistency because,

sin\theta_{1} \neq 1

What am I doing wrong?
 

Attachments

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Find the angle of incidence at the exit point of the ray. That should be greater than the critical angle so that internal reflection happen.

ehild
 
jegues said:

Homework Statement



See figure attached for problem statement.

Homework Equations





The Attempt at a Solution



I'm getting a disagreement in my equations so I must be doing something wrong.

We have two unknowns,

n_{1},\theta_{2}

My first equation,

n_{1}sin\theta_{1}=n_{2}sin\theta_{2}

my second equation (This is where I think I am misunderstanding something)

Since it is experiencing total internal reflection,

\theta_{2} = arcsin\frac{n_{1}}{n_{2}}  This should be:  90° ‒ θ2= arcsin(n1/n2),
which is: sin(90° ‒ θ2) = cos(θ2) = (n1/n2)


Or in other words,

sin\theta_{2} = \frac{n_{1}}{n_{2}}

If I plug this into my first equation I get an inconsistency because,

sin\theta_{1} \neq 1

What am I doing wrong?

See comment in red above.
 
SammyS said:
See comment in red above.

Thank you.
 

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