A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width ? what distance from the centre will the first maximum be located?

To solve this problem, we need to calculate the fringe width and the position of the first maximum in a double-slit interference experiment. The presence of mica and polystyrene strips introduces a phase difference due to their refractive indices and thicknesses. Let’s break this down step by step.

Understanding Fringe Width

The fringe width (β) in a double-slit experiment can be calculated using the formula:

β = λL / d

Where:

• λ = wavelength of light used (590 nm = 590 x 10^-9 m)
• L = distance from the slits to the screen (1 m)
• d = separation between the slits (0.12 cm = 0.0012 m)

Calculating Fringe Width

Substituting the values into the formula:

β = (590 x 10^-9 m) * (1 m) / (0.0012 m)

Calculating this gives:

β = 0.0004908333 m = 0.4908 mm

Phase Difference Due to the Strips

Next, we need to consider the phase difference introduced by the mica and polystyrene strips. The phase difference (Δφ) can be calculated using the formula:

Δφ = (2π/λ) * (n - 1) * t

Where:

• n = refractive index of the material
• t = thickness of the material

Calculating Phase Difference for Each Material

For mica:

Δφ_mica = (2π / (590 x 10^-9 m)) * (1.58 - 1) * (0.50 x 10^-3 m)

Calculating this gives:

Δφ_mica ≈ 2.67 radians

For polystyrene:

Δφ_polystyrene = (2π / (590 x 10^-9 m)) * (1.55 - 1) * (0.50 x 10^-3 m)

Calculating this gives:

Δφ_polystyrene ≈ 2.56 radians

Total Phase Difference

The total phase difference (Δφ_total) between the two strips is:

Δφ_total = Δφ_mica - Δφ_polystyrene

Substituting the values:

Δφ_total ≈ 2.67 - 2.56 = 0.11 radians

Finding the Position of the First Maximum

The position of the first maximum (y₁) can be calculated using the formula:

y₁ = (β/2) + (Δφ_total * L / (2π))

Substituting the values:

y₁ = (0.4908 mm / 2) + (0.11 * 1 / (2π))

Calculating this gives:

y₁ ≈ 0.2454 mm + 0.0175 m ≈ 0.2454 mm + 17.5 mm ≈ 17.7454 mm

Final Results

To summarize:

• The fringe width is approximately 0.4908 mm.
• The distance from the center to the first maximum is approximately 17.75 mm.

This analysis shows how the interference pattern is influenced by the materials used in the double-slit setup, and how we can calculate the resulting fringe width and positions of maxima effectively.