Waveguides and Numerical Aperture

AI Thread Summary
The discussion revolves around the calculations related to dielectric slab waveguides, specifically when cladding is replaced with air. The numerical aperture (NA) calculated is greater than 1, raising questions about its implications since the sine of an angle cannot exceed 1. This situation suggests that the critical angle should theoretically be zero, indicating potential issues with numerical aperture. Additionally, the role of cladding is examined, noting that a less optically dense cladding increases the acceptance angle of the waveguide. The conversation concludes with the notion that total internal reflection (TIR) at the core-air interface may lead to an optical cavity rather than a functioning waveguide.
GreeziakAttack
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Hey all,

First time posting on here hope this all goes well!

I just worked through a problem on dielectric slab waveguides with core and cladding and it was pretty straight forward finding critical angle, critical angle compliment, number of modes, numerical aperture, maximum acceptance angle etc.

Now the problem asks to run through all the calculations again sans the cladding (assume its air).

Here is where my questions start:

\ NA = \sqrt{n^{2}_{1}-n^{2}_{2}} is a number greater than 1 (1.25 to be exact)

Obviously here the sine of an angle cannot be greater than 1, so what does this say about the numerical aperture and the acceptance angle? It seems that this should not be so since I have reasonable values for critical angle and its complement. I would assume that the critical angle ought to be 0 for there to be issues with numerical aperture.

Also, in general what effect does the cladding have on the quality of the waveguide? Based on the numerical aperture it seems the less optically dense the cladding (or namely the ratio of the cladding to core indices) the higher the accepting angle of the guide.

EDIT:

Thinking about this more, the case where

\ \overline{θ} \geq θ_{c}

Means we will get TIR off the core air interface at the end of the waveguide, so essentially we have no waveguide but an optical cavity because all the rays remain inside the dielectric?
Any advice is greatly appreciated!

Thanks,
Keith
 
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Thinking about this more, the case where

\ \overline{θ} \geq θ_{c}

Means we will get TIR off the core air interface at the end of the waveguide, so essentially we have no waveguide but an optical cavity because all the rays remain inside the dielectric?
 
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