What is the Lowest Angle of θ1 for Total Internal Reflection?

[ParoXsitiC]

Homework Statement

Find the lowest angle of θ₁ given the apex angle is 60°. Air (n=1) is on the outside and inside (n=1.5)

ϕ is defined as 60 degrees.

Homework Equations

The Attempt at a Solution

θ_c = sin^-1($\frac{1.00}{1.50}$) = 41.81°

To my understanding, θ₁ must angle in such a way to make r (the angle of refraction) to be equal to the critical angle. At this point you will start having TIR.

They state that r = ϕ + θ_c
How? I am not seeing it.

Once I found r, I can just use snells law to find θ₁ - but I don't understand how to find r.

 

[PeterO]

Homework Statement

Find the lowest angle of θ₁ given the apex angle is 60°. Air (n=1) is on the outside and inside (n=1.5)


http://i.minus.com/1334731242/sW_NUCsWAiRgKlAAkKeqdw/ionULwjtTVYG1.png

ϕ is defined as 60 degrees.

Homework Equations

The Attempt at a Solution

θ_c = sin^-1($\frac{1.00}{1.50}$) = 41.81°

To my understanding, θ₁ must angle in such a way to make r (the angle of refraction) to be equal to the critical angle. At this point you will start having TIR.

They state that r = ϕ + θ_c
How? I am not seeing it.

Once I found r, I can just use snells law to find θ₁ - but I don't understand how to find r.

Need some sort of diagram so the position of this θ₁ is knows.

 

[ParoXsitiC]

Need some sort of diagram so the position of this θ₁ is knows.

I give one but perhaps its not showing for you since its minus.com, here it is on imgur

 

[PeterO]

I give one but perhaps its not showing for you since its minus.com, here it is on imgur

Image came through that time.

If you look at the top triangle - the one with the 60^o angle, the other two angles are (90 - r)^o and (90 - θ_c)^o

That means [(90-r) + ϕ + (90-θ_c)] = 180

so 180 -r + ϕ - θ_c = 180

or -r + ϕ - θ_c = 0 which means r = ϕ - θ_c

Notice that is slightly different to what was in your original solution. I suspect something was wrong.

 

[asteriatic]

Total internal reflection occurs when a light ray travels from a medium with a higher refractive index to a medium with a lower refractive index, and the angle of incidence is greater than the critical angle. In this case, the critical angle is 41.81°, which means that if the angle of incidence (θ1) is greater than 41.81°, total internal reflection will occur.

In order to find the lowest angle of θ1, we need to find the angle at which r (the angle of refraction) is equal to the critical angle. This can be done by using the formula r = ϕ + θc, where ϕ is the apex angle (60°) and θc is the critical angle (41.81°). Substituting these values, we get r = 60° + 41.81° = 101.81°.

This means that when the angle of incidence (θ1) is equal to 101.81° or greater, total internal reflection will occur. Therefore, the lowest angle of θ1 that will result in total internal reflection is 101.81°. Any angle between 41.81° and 101.81° will result in partial reflection and partial refraction.