Snells Law

  • Thread starter jegues
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  • #1
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,

[tex]n_{1},\theta_{2}[/tex]

My first equation,

[tex]n_{1}sin\theta_{1}=n_{2}sin\theta_{2}[/tex]

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

Since it is experiencing total internal reflection,

[tex]\theta_{2} = arcsin\frac{n_{1}}{n_{2}}[/tex]

Or in other words,

[tex]sin\theta_{2} = \frac{n_{1}}{n_{2}}[/tex]

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

[tex]sin\theta_{1} \neq 1[/tex]

What am I doing wrong?
 

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Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,915
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
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,603
1,190

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,

[tex]n_{1},\theta_{2}[/tex]

My first equation,

[tex]n_{1}sin\theta_{1}=n_{2}sin\theta_{2}[/tex]

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

Since it is experiencing total internal reflection,

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


Or in other words,

[tex]sin\theta_{2} = \frac{n_{1}}{n_{2}}[/tex]

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

[tex]sin\theta_{1} \neq 1[/tex]

What am I doing wrong?

See comment in red above.
 
  • #4
jegues
1,097
3

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