IAN LANG ELECTRONICS

Snell's law becomes important for calculating the critical angle because if we assume       2 to be 90 degrees the sine value is unity, and thus we can say  that the only factor is the refractive index. Now we transpose to find sin           then           (which is now the critical angle) and come up with the equation  

 

 =(arcsin (n   /n    ).

 

Let us consider what happens in graphic format:

Optoelectronics

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The drawing on the left shows a  boundary on the horizontal, the material on the lower half  having a higher refractive index than that of the upper. The green line shows the critical angle, the red a light ray approaching at an angle lesser than critical and the blue a ray approaching greater than critical. The normal is shown as the perpendicular axis.

The red ray in our example above would now go into the cladding, and not down the core. It may get reflected back to the core by the outers, but by this time much power will be lost. It may not, and then the whole transmission is lost. The blue ray on the other hand is mirrored back into the core at an angle by the boundary with very little if any loss. As it goes down again, it is mirrored once again, and rises. This effect continues something like the illustration below supplied by the US Navy (NEETS):

We will now imagine that the lower half is the core of an optical cable, the upper the cladding around it as in the drawing below which shows a fairly typical one.

 

 

 

 

 

 

 

 

 

As can be seen, the ray continues to be reflected at the same angle all down the length of the core with none being refracted by the cladding. This is the process of   total internal reflection. Of course if you had the light source emitting exactly on the axis and at the centre of the core, the ray would follow a straight line, but given the physical size of the cores this is difficult to achieve.