Refraction/Smallest curve of optical fibre.

[Berdi]

Homework Statement

"figure 3" (below) shows an optical fibre bent in a curve. The diameter of the fibre is 4mm. If light is not to escape, calculate the smallest radius of R of the curve round which the fibre can be bent. Assume that the fibre is surrounded by air (refractive index 1) and that the refractive index of glass is 1.5"

Homework Equations

n1
___ = SinC (critical angle)

n2

The Attempt at a Solution

Well, I've got c, the critical angle -

1
__ = 0.666... sin-1(o.666) = 41.81
1.5

But, I can seem to make a triangle with enough information to get R. I've tried splitting it up, but either get not enough information to do sine or cosine rules, or a non right angle triangle. Maybe I'm just missing something?

 

[mgb_phys]

Call the acute angle at the centre between the dashed and dotted lines A.
If the fibre makes a quadrant then the angle between the incoming ray and the dotted line is 90 therefore angle A is 90-C.
The dotted line is length R, the hypotonuse (dashed line) is R+4mm and the angle is (90-c)
So cos(90-C) = r/(r+4mm)

 

[Berdi]

So just rearrange that to give me R? Ill give it a shot.