Step-by-step Derivations

The problem asks for the number of identical units, n, a light ray passes through to achieve a total horizontal displacement l. Each unit consists of three layers: air, a medium with refractive index μ₁, and another medium with refractive index μ₂.

1. Analyze the path of the ray in a single unit using Snell's Law.

Let's trace the ray through one unit, calculating the angles of refraction at each interface. The initial angle of incidence at the first interface (Air to Medium 1) is given as θ = 60°.

• Interface 1 (Air → Medium 1):

  • Refractive index of air, $μ_air = 1$.
  • Refractive index of medium 1, μ₁ = √(3/2).
  • Angle of incidence, i₁ = θ = 60°.
  • Let the angle of refraction be r₁. According to Snell's Law: $μ_\text{air} \sin(i₁) = μ₁ \sin(r₁)$ $1 \cdot \sin(60°) = \sqrt{\frac{3}{2}} \cdot \sin(r₁)$ $\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{2}} \sin(r₁)$ $\sin(r₁) = \frac{\sqrt{3}}{2} \cdot \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ $r₁ = 45°$

• Interface 2 (Medium 1 → Medium 2):

  • The interfaces are parallel, so the angle of incidence is i₂ = r₁ = 45°.
  • Refractive index of medium 2, μ₂ = √3.
  • Let the angle of refraction be r₂. $μ₁ \sin(i₂) = μ₂ \sin(r₂)$ $\sqrt{\frac{3}{2}} \sin(45°) = \sqrt{3} \sin(r₂)$ $\sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} = \sqrt{3} \sin(r₂)$ $\frac{\sqrt{3}}{2} = \sqrt{3} \sin(r₂)$ $\sin(r₂) = \frac{1}{2}$ $r₂ = 30°$

• Interface 3 (Medium 2 → Air of the next unit):

  • The angle of incidence is i₃ = r₂ = 30°.
  • Let the angle of emergence be r₃. $μ₂ \sin(i₃) = μ_\text{air} \sin(r₃)$ $\sqrt{3} \sin(30°) = 1 \cdot \sin(r₃)$ $\sqrt{3} \cdot \frac{1}{2} = \sin(r₃)$ $\sin(r₃) = \frac{\sqrt{3}}{2}$ $r₃ = 60°$

Since the angle of emergence into the air layer of the next unit is 60°, which is the same as the initial angle of incidence, the path of the ray will be identical in each subsequent unit.

2. Calculate the horizontal displacement in one unit (Δx).

The total horizontal displacement in one unit is the sum of the displacements in each of the three layers.

• Displacement in Layer 1 (Air):

  • Height h = 1/3 cm. Angle with normal is 60°.
  • $x_\text{air} = h \tan(60°) = \frac{1}{3} \cdot \sqrt{3} = \frac{1}{\sqrt{3}}$ cm.

• Displacement in Layer 2 (Medium 1):

  • Thickness d = (√3 - 1)/2 cm. Angle with normal is r₁ = 45°.
  • $x_\text{med1} = d \tan(r₁) = d \tan(45°) = \frac{\sqrt{3}-1}{2} \cdot 1 = \frac{\sqrt{3}-1}{2}$ cm.

• Displacement in Layer 3 (Medium 2):

  • Thickness d = (√3 - 1)/2 cm. Angle with normal is r₂ = 30°.
  • $x_\text{med2} = d \tan(r₂) = d \tan(30°) = \frac{\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3}} = \frac{\sqrt{3}-1}{2\sqrt{3}}$ cm.

Total displacement per unit Δx: $Δx = x_\text{air} + x_\text{med1} + x_\text{med2}$ $Δx = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}-1}{2} + \frac{\sqrt{3}-1}{2\sqrt{3}}$ To simplify the sum, let's find a common denominator, which is 6: $Δx = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}-1}{2} + \frac{3-\sqrt{3}}{6}$ $Δx = \frac{2\sqrt{3}}{6} + \frac{3(\sqrt{3}-1)}{6} + \frac{3-\sqrt{3}}{6}$ $Δx = \frac{2\sqrt{3} + 3\sqrt{3} - 3 + 3 - \sqrt{3}}{6}$ $Δx = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}} \text{ cm}$

3. Determine the number of units, n.

The total horizontal displacement l after passing through n identical units is n times the displacement per unit Δx. $l = n \cdot Δx$ We are given l = 8/√3 cm. $\frac{8}{\sqrt{3}} = n \cdot \frac{2}{\sqrt{3}}$ $8 = 2n$ $n = 4$

The value of n is 4.