In a Young's double-slit experiment with light of wavelength 500 nm, a thin transparent sheet of refractive index 1.5 and thickness 8 mum covers one…

A sheet adds an extra optical path $(\mu-1)t$. For the first sheet: $0.5\times8000\ \mathrm{nm} = 4000\ \mathrm{nm}$; for the second: $0.4\times5000\ \mathrm{nm} = 2000\ \mathrm{nm}$. The sheets are on opposite slits, so the net path difference is $4000-2000 = 2000\ \mathrm{nm}$. The number of fringes shifted is $\dfrac{2000}{500} = 4$. The zero-path-difference point moves toward the slit with the larger added optical path, i.e. the one covered by the $\mu=1.5$ sheet. Adding the two paths (giving $12$) ignores that they oppose.