Optical Path And Thin Films

Placing a transparent slab before one slit adds extra optical path $(n-1)t$ to that beam, displacing the whole fringe pattern toward the covered slit. The number of fringes by which the pattern shifts is $N=\frac{(n-1)t}{\lambda }$, since each fringe corresponds to one wavelength of path difference. Substituting $n=1.5$, $t=10\times 10^{-6}$ m and $\lambda =500\times 10^{-9}$ m gives $N=\frac{(1.5-1)\times 10\times 10^{-6}}{500\times 10^{-9}}=\frac{0.5\times 10^{-5}}{5\times 10^{-7}}=\frac{5\times 10^{-6}}{5\times 10^{-7}}=10$. The key insight is that interference responds to optical path, not geometric path: inside the slab the light traverses the same physical thickness but, travelling slower, accumulates as many extra wavelengths as if it had crossed an additional vacuum length $(n-1)t$. Only this excess over the path the beam would have had in air shifts the fringes, which is why the slab thickness never appears without the $(n-1)$ factor. Because the shift came out an exact integer, the central bright fringe lands precisely where the tenth fringe formerly sat. The value 5 is wrong because it omits the factor $(n-1)$ and uses $t/\lambda$ scaled incorrectly. The value 20 is wrong as it mistakenly uses the full index $n$ rather than $n-1$. The value 15 has no consistent derivation. This NCERT optical-path argument explains fringe displacement by thin films. A check confirms only the excess path $(n-1)t$ matters, not the full slab thickness, giving a clean integer shift of 10.