What is the Minimum Refractive Index for Total Internal Reflection?

[jegues]

Homework Statement

See figure attached for problem statement.

Homework Equations

The Attempt at a Solution

Using Snell's Law,

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

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

Where,

\theta_{1} = 30^{o}, \theta_{2} = 90^{o}

It gives me, n_{1} = 2 but the answer is a minimum of 1.15.

What did I do wrong/misunderstand?

 

[tiny-tim]

hi jegues!

it's not 30°, it's 60°

 

[jegues]

hi jegues!

it's not 30°, it's 60°

Yes I figured that much but I don't understand why.

I'm looking at the triangle and the angle at the bottom right should be 90-60 = 30, that's where the incident angle is isn't it?

 

[tiny-tim]

hi jegues!

… the angle at the bottom right should be 90-60 = 30, that's where the incident angle is isn't it?

nooo … the angles of incidence and refraction are always from the normal

 

[jegues]

hi jegues!


nooo … the angles of incidence and refraction are always from the normal

So a line perpendicular to the surface it's hitting, right?

 

[tiny-tim]

s'right!