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# 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 $420nm$ and $630nm$. The minimum thickness of the soap solution is$420nm$$500nm$$450nm$$490nm$

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Solution

## The correct option is C $450nm$Step 1: Given Data:Refractive index of soap solution, ${\mu }_{s}=1.4$Refractive index of glass, ${\mu }_{g}=1.5$Two adjacent reflection maxima are observed at two wavelengths, ${\lambda }_{1}=420nm\phantom{\rule{0ex}{0ex}}{\lambda }_{2}=630nm$Let the thickness of the soap solution is $t$.Step 2: We have to find the minimum thickness of the soap solution is,For reflection at the air-soap solution interface, the phase difference is$\pi$.For reflection at the interface of soap solution to glass also, there will be a phase difference of $\pi$.The condition for maximum intensity:$2\mu t=n\lambda$ (where, $\mu$is the refractive index and $n$ is order) Step 3: For order $n$:As we know, the formula of refractive index is ${n}_{1}{\lambda }_{1}={n}_{2}{\lambda }_{2}$For n, $n{\lambda }_{1}=\left(n-1\right){\lambda }_{2}$$⇒$ $n×420=\left(n-1\right)630$$⇒$ $n×420=n×630-630$$⇒$$n×420-n×630=-630$$⇒$ $n\left(420-630\right)=-630$$⇒$ $n\left(-210\right)=-630$$⇒$ $210n=630$$⇒$ $n=\frac{630}{210}$$⇒$ $n=3$Step 4: Maximum order, where they coincide: $2×1.4×t=3×420$$⇒$ $t=\frac{2×1.4}{3×420}$$⇒$ $t=450nm$Hence, the correct option is (C).  Suggest Corrections  0      Explore more