A thin oil film of refractive index 1.2 float | vedclass

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Question

(B) For normal incidence,the path difference between the ray reflected from the top surface and the ray reflected from the bottom surface is $\Delta x

Vedclass · Accepted Answer

$2 	imes 10^{-7} \ m$

Vedclass · Answer

(B) For normal incidence,the path difference between the ray reflected from the top surface and the ray reflected from the bottom surface is $\Delta x = 2 \mu_1 t$.Since the refractive index of oil $(\mu_1 = 1.2)$ is less than that of water $(\mu_2 = 1.33)$,there is no phase change at the bottom interface,but there is a phase change of $\pi$ at the top interface (reflection from a denser medium).For destructive interference (dark appearance),the condition is $2 \mu_1 t = n \lambda$.For constructive interference (bright appearance),the condition is $2 \mu_1 t = (n + \frac{1}{2}) \lambda$.To change from dark to bright,the change in path difference must be $\Delta(2 \mu_1 t) = 2 \mu_1 \Delta t = \frac{\lambda}{2}$.$\Delta t = \frac{\lambda}{4 \mu_1} = \frac{9.6 	imes 10^{-7}}{4 	imes 1.2} = \frac{9.6 	imes 10^{-7}}{4.8} = 2 	imes 10^{-7} \ m$.

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