In Young's Double Slit Experiment (YDSE), a glass slab of thickness 8 μm is introduced in front of one slit. The central maximum shifts to the position of the 4th minimum.

Given:

• Thickness of glass slab, t = 8 μm
• Wavelength, λ = 500 nm

Find the refractive index of the glass slab.

When a thin glass slab is introduced in front of one slit in YDSE, an additional optical path difference is introduced.

Additional path difference:

Δ = (μ − 1)t

The central bright fringe shifts to the position of the 4th minimum.

For the nth minimum:

Path difference = (2n−1)λ/2

For 4th minimum:

Δ = 7λ/2

Substituting:

(μ−1)t = 7λ/2

(μ−1)(8×10⁻⁶) = 7(500×10⁻⁹)/2

μ−1 = 3.5×500×10⁻⁹ / 8×10⁻⁶

μ−1 = 0.21875

μ ≈ 1.22

Final Answer

Refractive Index = 1.22