Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.33/2.42) ≈ 36.9°.

To determine if total internal reflection occurs, we first find the critical angle using sin(θc) = 1/n = 1/1.5, which gives θc ≈ 41.8°. Since 60° > 41.8°, total internal reflection will not occur.

Since the angle of incidence (60°) is greater than the critical angle (approximately 41.8° for glass to air), total internal reflection occurs.

Critical angle θc = sin⁻¹(1/2.42) ≈ 24.4°.

Critical angle θc = sin^(-1)(1/n) = sin^(-1)(1/2.42) ≈ 24.4°.

For total internal reflection to occur in a fiber optic cable, the core must have a higher refractive index than the cladding.

Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.5/1.6) ≈ 38.7°.

Critical angle θc = sin⁻¹(n2/n1) = sin⁻¹(1.4/1.5) ≈ 42.0°.

For total internal reflection, the core must have a higher refractive index than the cladding, so it must be greater than 1.45.

The cladding has a lower refractive index than the core, ensuring that light is kept within the core through total internal reflection.

The cladding has a lower refractive index than the core, ensuring that light remains within the core by total internal reflection.

Since the angle of incidence is less than the critical angle (θc = sin⁻¹(1/1.5) ≈ 41.8°), it will not undergo total internal reflection.

Critical angle θc = sin⁻¹(1.00/2.42) ≈ 24.4°.

The critical angle for this scenario is approximately 38.7°. Since 50° is greater than the critical angle, total internal reflection occurs.

Since 50° is greater than the critical angle (θc ≈ 41.8°), total internal reflection occurs.

Calculate critical angle: θc = sin^(-1)(1/n) = sin^(-1)(1/1.6) ≈ 38.7°. Since 50° > 38.7°, total internal reflection occurs.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 2.0, θ1 = 45°, and n2 = 1.0 (air). Thus, 2.0 * sin(45°) = 1.0 * sin(θ2) leads to sin(θ2) = √2, which gives θ2 = 90°.

Using Snell's law, sin(θc) = n2/n1 = 1.00/2.42. Thus, θc ≈ 24.4°.

Using sin(θc) = n2/n1, where n1 = 2.42 (diamond) and n2 = 1.0 (air), we find sin(θc) = 1.0/2.42, leading to θc ≈ 24.4°.

Since sin(30°) < sin(θc) where θc = sin⁻¹(1/1.5) ≈ 41.8°, it will not undergo total internal reflection.

Since 45° < θc ≈ 48.6°, the ray will partially reflect and refract.

Using Snell's law, n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, θ1 = 30°, n2 = 1.0. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, giving θ2 ≈ 60°.

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), 1 * sin(60°) = 1.33 * sin(θ2) => sin(θ2) = (sin(60°)/1.33) ≈ 0.577, θ2 ≈ 35.0°.

Since the angle of incidence (70°) is greater than the critical angle (approximately 41.8°), total internal reflection occurs.

The critical angle for diamond to air is sin⁻¹(1/2.42) ≈ 24.4°. Since 70° > 24.4°, total internal reflection will occur.

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), 1.5 * sin(45°) = 1 * sin(θ2) => sin(θ2) = 1.5 * 0.7071 = 1.0607, which is not possible, hence total internal reflection occurs.

Since 70° > θc = sin⁻¹(1/1.5) ≈ 41.8°, total internal reflection will occur.

Since the critical angle for glass to air is sin⁻¹(1.00/1.5) ≈ 41.8°, and 60° > 41.8°, total internal reflection occurs.

Using Snell's law: n1 * sin(θ1) = n2 * sin(θ2). Here, n1 = 1.5, n2 = 1.0, and θ1 = 30°. Thus, sin(θ2) = (1.5 * sin(30°))/1.0 = 0.75, leading to θ2 ≈ 48.6°.

Since the angle of incidence (70°) is greater than the critical angle (approximately 41.8° for glass to air), total internal reflection occurs.