A light ray of wavelength 5000A° travelling through a medium of refractive index 1.5 gets incident at an angle of 30° on another medium of refractive index 1.25 and thickness 0.2mm. Calculate the minimum phase difference and path difference between the rays coming back into the initial medium from the two faces of the second medium....

To tackle the problem of calculating the minimum phase difference and path difference for a light ray traveling through two different media, we need to break it down into manageable steps. Let's start by understanding the scenario and the relevant concepts involved.

Understanding the Setup

We have a light ray with a wavelength of 5000 Å (angstroms), which is equivalent to 5000 x 10^-10 meters. The ray is incident at an angle of 30° on a medium with a refractive index of 1.25, which has a thickness of 0.2 mm (or 0.2 x 10^-3 meters). The initial medium has a refractive index of 1.5.

Calculating the Path Lengths

When the light ray enters the second medium, it will experience a change in speed and direction due to refraction. To find the path difference, we first need to calculate the distance the light travels in the second medium.

• The angle of refraction can be found using Snell's Law: n₁ sin(θ₁) = n₂ sin(θ₂).
• Here, n₁ = 1.5, θ₁ = 30°, and n₂ = 1.25.

Calculating sin(30°) gives us 0.5. Plugging in the values:

1.5 * 0.5 = 1.25 * sin(θ₂)

Solving for sin(θ₂):

sin(θ₂) = (1.5 * 0.5) / 1.25 = 0.6

Now, we can find θ₂ using the inverse sine function:

θ₂ = sin^-1(0.6) ≈ 36.87°.

Calculating the Path Length in the Second Medium

Next, we need to find the actual path length the light travels within the second medium. The thickness of the medium is 0.2 mm, and we can find the length of the path using trigonometry:

Path length (L) = thickness / cos(θ₂)

Calculating cos(36.87°) gives approximately 0.8:

L = (0.2 x 10^-3) / 0.8 = 0.25 x 10^-3 meters.

Calculating the Path Difference

The path difference (ΔL) between the rays coming back into the initial medium from the two faces of the second medium can be calculated as:

ΔL = 2 * L = 2 * (0.25 x 10^-3) = 0.5 x 10^-3 meters.

Calculating the Phase Difference

The phase difference (Δφ) can be calculated using the formula:

Δφ = (2π/λ) * ΔL

Substituting the values:

λ = 5000 Å = 5000 x 10^-10 m

Δφ = (2π / (5000 x 10^-10)) * (0.5 x 10^-3)

Calculating this gives:

Δφ ≈ (2π / 5000 x 10^-10) * (0.5 x 10^-3) ≈ 6.2832 * (0.5 / 5000) x 10⁷ ≈ 6.2832 * 10^-4 radians.

Final Results

To summarize, the minimum path difference between the rays returning to the initial medium is approximately 0.5 mm, and the corresponding phase difference is about 6.2832 x 10^-4 radians. This analysis illustrates how light behaves when transitioning between different media, showcasing the principles of refraction and wave behavior.