Q.When light is incident on a soap film of thickness $5\times10^{-5}$ cm, the wavelength of light reflected maximum in the visible region is 5320 Å. The refractive index of the film will be,

A thin transparent film of refractive index $\mu$ and thickness $d$, illuminated by a parallel beam at (near-)normal incidence, splits the light at its upper surface into a reflected part and a refracted part; the refracted part is itself partly reflected at the lower surface and re-emerges through the top, so both the reflected light and the transmitted light from a thin film consist of two (or more) coherent beams that can interfere, giving the film a characteristic colour (as seen in soap bubbles and oil slicks). For the extra optical path $2\mu d$ travelled inside the film by the twice-refracted wave (approximated for near-normal incidence as simply twice the film thickness, weighted by the refractive index), the transmitted light interferes constructively when $2\mu d=n\lambda$ and destructively when $2\mu d=(2n-1)\lambda/2$. The reflected light, however, picks up one extra phase shift of $\pi$ (equivalent to an extra path of $\lambda/2$) whenever light reflects off a surface while travelling from a rarer into a denser medium -- so the reflected-light conditions are exactly reversed relative to the transmitted-light ones: constructive interference for reflected light requires $2\mu d+\lambda/2=n\lambda$, i.e. $2\mu d=(2n-1)\lambda/2$, and destructive interference requires $2\mu d=n\lambda$. (For a general, not-quite-normal angle of incidence with internal refraction angle $r$, the path term $2\mu d$ is replaced everywhere by $2\mu d\cos r$.) These conditions are exploited to design thin anti-reflection coatings (destruct …