# Given a 45 45 90 prism with index of 1.5, immersed in air. The hypotenuse acts as the reflecting face by TIR. A ray of light enters horizont

Given a 45 45 90 prism with index of 1.5, immersed in air. The hypotenuse acts as the reflecting face by TIR. A ray of light enters horizontal (perpendicular to one of the non- hypotenuse faces) hitting the hypotenuse at 45º and is reflected. If the original ray is now the center of a converging bundle of rays, converging toward the original intersection point on the hypotenuse, what full convergence angle (in air) is allowed, if TIR at the hypotenuse is to be preserved for the whole bundle of rays?

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83.6°

Explanation:

For the ray to be totally internally reflected, at the boundary, the angle of refraction is 90. Using the law of refraction where

n₁sinθ₁ = n₂sinθ₂ where n₁ = refractive index of prism = 1.5, θ₁ = critical angle in prism, n₂ = refractive index of air = 1 and θ₂ = refractive angle = 90°.

So, substituting these values into the equation,

n₁sinθ₁ = n₂sinθ₂

1.5 × sinθ₁ = 1 × sin90

1.5 × sinθ₁  = 1

sinθ₁ = 1/1.5

sinθ₁ = 0.6667

θ₁  = sin*(0.6667)

θ₁  = 41.8°

So, for total internal reflection, an incidence angle of 41.8° is required. So, a full convergence angle of 2 × 41.8° = 83.6° is required for the whole bundle of rays.