A fish inside water (μ = 4/3) sees the world above

Solution

Given:

• The fish is in water with refractive index μ = 4/3
• The fish looks up at the air-water interface

When a fish looks up at the water-air interface, light rays from above enter the water. However, there's a critical angle beyond which light from water cannot pass into air (total internal reflection). This creates a circular "window" through which the fish can see the outside world. The critical angle θc occurs when the refracted angle in air is 90°. Using Snell's law: μ₁ sin θ₁ = μ₂ sin θ₂ Where:

• μ₁ = 4/3 (water)
• μ₂ = 1 (air)
• θ₁ = θc (critical angle in water)
• θ₂ = 90° (refracted angle in air)

Therefore: (4/3) sin θc = 1 × sin 90° (4/3) sin θc = 1 sin θc = 3/4 θc = sin^-1(3/4) ≈ 48.6° The circular window the fish sees is formed by the cone of angles less than the critical angle. From the fish's perspective, this creates a circular window with an angular diameter of 2θc. Angular diameter = 2θc = 2 × 48.6° ≈ 97.2°