An optical fibre is made of flint glass of index 1.66 clad with crown glass of index 1.52. What is the critical angle at the core to cladding boundary?

1. Light travels inside the flint glass core, so n1 = 1.66, and the crown glass cladding gives n2 = 1.52. 2. The condition n1 greater than n2 is met, so a critical angle exists. 3. Substituting into theta_c = inverse sine of (n2 divided by n1) gives inverse sine of (1.52 divided by 1.66). 4. The bracket is 0.916, so theta_c = inverse sine of 0.916 = 66.3 degrees. 5. Only rays striking the wall at more than 66.3 degrees stay in the core, which is why the two indices are kept close in practice. 6. Inverting the ratio gives a sine greater than one, which has no solution, so any answer built that way is meaningless. _Source: OpenStax College Physics (CC BY 4.0), Ch 25 "Geometric Optics", section 25.4 Total Internal Reflection_