In YDSE, a plate of thickness t = 4/3 μm and refractive index 1.5 is placed before one slit. Temperature coefficient of refractive index is 3 × 10⁻³ °C⁻¹. Incident wavelength is 6000 Å. The temperature rises by 100 °C without plate expansion. Find number of fringes crossing a fixed screen point.
Physics · Mixed Chapter
JEE Main Practice

Question

In YDSE, a plate of thickness t = 4/3 μm and refractive index 1.5 is placed before one slit. Temperature coefficient of refractive index is 3 × 10⁻³ °C⁻¹. Incident wavelength is 6000 Å. The temperature rises by 100 °C without plate expansion. Find number of fringes crossing a fixed screen point.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Index change is Δμ = μαΔT, where α = 3 × 10⁻³ °C⁻¹. Step 2 — Δμ = 1.5 × 3 × 10⁻³ × 100 = 0.45. Step 3 — Additional optical path difference is Δ = t Δμ = (4/3 × 10⁻⁶) × 0.45 = 6 × 10⁻⁷ m. Step 4 — Wavelength is 6000 Å = 6 × 10⁻⁷ m. Number of fringes crossing point = Δ/λ = 1.