White light reflected from a soap film (Refra | vedclass

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(C) For a thin film,the condition for constructive interference (maxima) in reflected light is $2 \mu t \cos r = (n + 1/2) \lambda_1$,where $n$ is an

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$3$

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(C) For a thin film,the condition for constructive interference (maxima) in reflected light is $2 \mu t \cos r = (n + 1/2) \lambda_1$,where $n$ is an integer.For destructive interference (minima),the condition is $2 \mu t \cos r = m \lambda_2$,where $m$ is an integer.Given $\lambda_1 = 600 \ nm$ and $\lambda_2 = 450 \ nm$. Assuming normal incidence $(\cos r = 1)$:$2 \mu t = (n + 1/2) \lambda_1 = (2n + 1) \frac{\lambda_1}{2} = (2n + 1) 	imes 300 \ nm$$2 \mu t = m \lambda_2 = m 	imes 450 \ nm$Equating the two: $(2n + 1) 	imes 300 = m 	imes 450 \implies (2n + 1) 	imes 2 = 3m \implies 4n + 2 = 3m$.For the smallest thickness with no minimum in between,we test $n=1$: $4(1) + 2 = 6 = 3m \implies m = 2$.Substituting $n=1$ into the maxima condition:$2 	imes 1.5 	imes t = (1 + 0.5) 	imes 600 \ nm$$3t = 1.5 	imes 600 \ nm = 900 \ nm$$t = 300 \ nm = 3 	imes 10^{-7} \ m$.Thus,the thickness is $3$ units.

White light reflected from a soap film (Refractive Index $\mu = 1.5$) has a maxima at $600 \ nm$ and a minima at $450 \ nm$ with no minimum in between. Then the thickness of the film in units of $10^{-7} \ m$ is:

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