The figure illustrates a basic optical fiber consisting of a plastic core (\(n_1 = 1.60\)) surrounded by a plastic sheath (\(n_2 = 1.50\)). A light ray enters one end of the fiber at an angle \(\theta\). This ray is meant to undergo total internal reflection at point A, where it meets the core-sheath boundary, ensuring no light is lost through that boundary. The question posed is: What is the maximum value of \(\theta\) that permits total internal reflection at point A?Below the question, there is an input field labeled "Number" for entering the calculated value of \(\theta\), followed by a unit selection dropdown menu. Lastly, a note mentions that the absolute tolerance for the answer is \(\pm 0.1\).**Diagram Explanation:**The diagram shows a cylindrical optical fiber. The core (light blue) has a refractive index of \(n_1 = 1.60\), while the yellow outer sheath has a refractive index of \(n_2 = 1.50\). A red line represents the path of the light ray entering the fiber at an angle \(\theta\). At point A, inside the core, the ray reflects off the boundary into the core, illustrating the total internal reflection process.

Given: The refractive index of plastic core n1 = 1.60. The refractive index of plastic core n2 = 1.50.The equation for Snell's law for total internal reflection at A can be given by, n1sinθ=n2sinθ1 Here, θ and θ1 are the incident and refracted angle with the fibre. Substituting the required values in the above equation to obtain the maximum angle of θ. 1.6×sinθ=1.5×sinθ1sinθ=0.93sinθ1 Sinθ will be maximum when the value of Sinθ1 will be maximum that is 1, now the equation can be given by, sinθ=0.93×1θ=sin-10.93θ=68.43∘ Therefore, the maximum value of θ is 68.43o.

Answered: The figure depicts a simplistic optical…