Trapped light

A figure to substantiate the solution would have been nice. But bear with it.

Let us consider the time when the light ray has entered the cylinder and is incident on the glass-coating interface. The angle of incidence is $\alpha$ . So, $n_{glass} \cdot \sin \alpha = n_{coating} \cdot \sin 90^\circ$

Thus the angle of refraction in air-glass interface is $90^\circ -\alpha$ .

So, $\theta$ and $\alpha$ are related by,

$n_{glass} \cdot \cos \alpha = n_{air} \cdot \sin \theta$

which gives $\sin \theta = \sqrt{(n_{glass}^2 - n_{coating}^2)}$ , i.e. $\theta$ is around 20.12.

Call the angle of refraction from the air to the glass $\alpha$ . Then by Snell's Law we have:

$n_{air}\sin \theta = n_{glass}\sin \alpha$

Call the angle of incidence from the glass to the coating $\beta$ , and the angle of refraction from the glass to the coating $\gamma$ . Then we have:

$n_{glass}\sin \beta = n_{coating}\sin \gamma$

In order for the light to be not be transmitted through the coating, we want $\gamma = 90^\circ$ . We also note that $\beta = 90^\circ - \alpha$ , since the surface runs parallel to the axis. So we have:

$n_{glass}\sin (90^\circ - \alpha) = n_{coating}\sin 90^\circ$

Substituting values, we solve for $\alpha = 13.2615^\circ$ .

Then we have:

$n_{air}\sin \theta = n_{glass}\sin 13.2615^\circ$

$\theta = 20.1264^\circ$

Applying the law of refraction, we deduce that the angle of incidence of light passing from glass fiber to the coating must be larger than $r_{limit} = arcsin \frac{n_c}{n_g} = 76.7385^\circ$

$\rightarrow i\leq 90^\circ - r_{limit} = 13.2615^\circ$ (i is the angle of refraction when light pass from air to glass)

$\rightarrow \theta \leq arcsin(n_g\sin i ) \Rightarrow \theta \leq 20.126^\circ$

1. theta: angle of incidence, alpha: angle of refraction between air and the glass that

---> sin(theta) = (n glass/n air)*sin(alpha)

1. 90 deg - alpha: critical angle between the glass and the coating of the fiber

---> sin(90-alpha) = cos(alpha) = (n coating/n glass)

1. sin(alpha) = sqrt(1-(cos(alpha))^2)

2. theta = arcsin[(n glass/n air) sqrt(1-(n_coating/n_glass)^2)] = arcsin[(1.50) sqrt(1-(1.46/1.50)^2)] = arcsin(0.34409) = 20.126 deg

In order to a total reflexion the Snell's Law must be analysed: n1sina=n2sinb (. N1 and n2 are the refective index of the medias, and "a" and "b" the angles of incidence and refration). In that case, sin(b) must be higher or equal to 1. It is known that doesn't exist such an angle which sine>1, what really happens is that when the caculeted angle turns into 1, the light rays won't pass for the other media, occuring the total reflection. In the problem, the angle which the light makes with axis of the fiber at the glass is relatated with angle of incidence \alpha

The entering light refracts according to the law $n_{air} \sin(\alpha) = n_{glass} \sin(\beta).$

When it hits the coating-glass boundary, it refracts according to the equation: $n_{glass} \cos(\beta) = n_{coating} \sin(\frac{\pi}{2}) = n_{coating}.$

Therefore, the entering angle should be $\sin(\alpha) = \frac{n_{glass}}{n_{air}} \times \sqrt{1 - (\frac{n_{coating}}{n_{air}})^2}.$

Solving for $\alpha$ we get $20.126^\circ$ .

We have critical angle of this problem

$\sin \theta _c=\cos \theta_1=\frac {n_2}{n_1}....(1)$ ,

Use snell laws we get:

$n_{air}\cdot \sin \theta =n_1\cdot \sin \theta _1$ ,

$\sin \theta _1=\frac{\sin\theta}{n_1}....(2)$ ,

Use (...1) and (...2), we get: $\left (\frac{n_2}{n_1} \right )^2+\left (\frac{\sin \theta }{n_1} \right )^2=1$ , $\theta =\arcsin\sqrt{n_1^2-n_2^2}=\arcsin\sqrt{1.50^2-1.46^2}=20.13^{\circ}$