Q. A thin film of soap solution $(\mu_{s} = 1.4)$ lies on the top of a glass plate $(\mu_{g} = 1.5)$. When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and 630 $nm$. The minimum thickness of the soap solution are
A) $420\ nm$
B) $450\ nm$
C) $630\ nm$
D) $1260\ nm$
Explanation
(b) For reflection at the air-soap solution interface, the phase difference is $\pi$
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For reflection at the interface of soap solution to glass also there will be a phase difference of $\pi$
$\therefore$ The condition for max. intensity $= 2\mu t = n\lambda$
For $n,n\lambda_{1} = (n - 1)\lambda_{2}$
$n\ 420nm = (n - 1)630\ nm$
$n(630 - 420) = 630,\ \therefore n(210) = 630\ nm$
$\therefore n = \frac{630}{210} = 3$
This is the maximum order where they coincide
$2 \times 1.4 \times t = 3 \times 420 \Rightarrow t = \frac{3 \times 420}{2 \times 1.40} = 450\ nm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.