# Snells Law

jegues

## Homework Statement

See figure attached for problem statement.

## 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

• SNELLQ.JPG
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## Answers and Replies

Homework Helper
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

Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

See figure attached for problem statement.

## 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.

jegues
See comment in red above.

Thank you.