Should optical cables be water tight? Geometric optics

1. The problem statement, all variables and given/known data

Explain the physical principle of total internal reflection used by optical cables. Calculate the critical angle of incidence that corresponds to a refracted angle θair = 90
Next, calculate the critical angle for a bare glass fiber submerged in water nH2O = 1.33.

Should optical cables be water tight?

2. Relevant equations

n1sinθ1=n2sinθ2

3. The attempt at a solution
2. As the angle of light passing from glass to air increases with respect to the normal line, the refracted light bends further away from the normal line. At the critical angle, the refracted light will not pass through the glass but will travel 90° along the surface of the glass. An optical cable works because light travels through the core of the cable which has a higher index of refraction than the cladding surrounding the core. The angle of light is always greater than the critical angle so the light is always reflected from the cladding. This allows the light to travel great distances.
Θcr of θair = 90° is sin-1 (nr/ni) = sin-1 (1/1.5) = 41.8°
Θcr of θH2O = 90° is sin-1 (nr/ni) = sin-1 (1.33/1.5) = 62.457°

I’m just having trouble with the last question. I think that optical cables should be water tight because the critical angle for glass to water is greater than that of glass to air. This means that the light will have total internal reflection at angles greater than 41.8 which is easier to achieve than angles greater than 62.457.

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