Total Internal Reflection Confusion

[jegues]

Homework Statement

See figure attached.

Homework Equations

The Attempt at a Solution

Initially,

[tex]\theta_{c} = 60^{o}[/tex]

[tex]n_{glass}sin\theta_{c} = n_{air}[/tex]

So,

[tex]n_{glass} = \frac{n_{air}}{sin\theta_{c}}[/tex]

Now I would think that the incident angle from glass to air would be able to get LARGER and still remain critical because it would be less "head on" to the incident surface and thus less likely to refract through.

So if we assume, [tex]\theta_{c}[/tex] can get larger and still remain critical then if we take and angle say,

[tex]\theta_{c} = 85^{o}[/tex]

certainly,

[tex]n_{glass} = \frac{n_{air}}{sin\theta_{c}} < 1.15[/tex]

because the value of the denominator is getting larger.

However they give c) as the correct answer. What am I misunderstanding?

 

[Doc Al]

Now I would think that the incident angle from glass to air would be able to get LARGER and still remain critical because it would be less "head on" to the incident surface and thus less likely to refract through.

Think of it this way. Since the angle shown demonstrates total internal reflection, the critical angle must equal that or less. So that defines a minimum value for the index of refraction of the glass.

So if we assume, [tex]\theta_{c}[/tex] can get larger and still remain critical then if we take and angle say,

[tex]\theta_{c} = 85^{o}[/tex]

Surely a larger angle would still demonstrate total internal reflection, but that larger angle would not be the critical angle.

 

[jegues]

Think of it this way. Since the angle shown demonstrates total internal reflection, the critical angle must equal that or less. So that defines a minimum value for the index of refraction of the glass.


Surely a larger angle would still demonstrate total internal reflection, but that larger angle would not be the critical angle.

This is much more clear, thank you.