Question

A soap film of thickness $0.3\mu m$ appears dark when seen by the reflected light of wavelength $580nm$. What is the index of reflection of the soap solution if it is known to be in between $1.3$ and $1.5$?

Answer 1

Given that,

Soap film thickness, $d=0.4\times {10}^{-6}m$

Wavelength, $\lambda =580\times {10}^{-9}m$

Soap film appear dark (minimum interference of light)

For, minimum illuminiation reflection of light, $2\mu d=n\lambda$

At $n=2$

$\Rightarrow \mu =\frac{n\lambda }{2d}=\frac{2\times 580\times {10}^{-9}}{2\times 0.4\times {10}^{-6}}=1.45$

Refractive index film, $\mu =1.45$