Finding n₁ when light undergoes total internal reflection

[jegues]

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?

 

[ehild]

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

 

[SammyS]

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.

 

[jegues]

See comment in red above.

Thank you.