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(B) For a thin film of refractive index $n_f$ on a substrate with refractive index $n_s$,where $n_s > n_f > n_{air}$,the condition for constructive in

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$450$

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(B) For a thin film of refractive index $n_f$ on a substrate with refractive index $n_s$,where $n_s > n_f > n_{air}$,the condition for constructive interference (reflection maxima) at normal incidence is given by $2 n_f t = (m + 1/2) \lambda$,where $m$ is an integer.Since we are given two adjacent maxima at $\lambda_1 = 420 \ nm$ and $\lambda_2 = 630 \ nm$,let the order for $\lambda_1$ be $(m+1)$ and for $\lambda_2$ be $m$.$2 n_f t = (m + 1 + 1/2) \lambda_1 = (m + 1/2) \lambda_2$$(m + 3/2) 	imes 420 = (m + 1/2) 	imes 630$$(2m + 3) 	imes 210 = (2m + 1) 	imes 315$$(2m + 3) 	imes 2 = (2m + 1) 	imes 3$$4m + 6 = 6m + 3$$2m = 3 \implies m = 1.5$ (This suggests the condition $2 n_f t = m \lambda$ applies if phase shifts occur at both surfaces).Actually,for $n_{air} $m \lambda_1 = (m-1) \lambda_2$$m 	imes 420 = (m-1) 	imes 630$$2m = 3m - 3 \implies m = 3$.$2 n_f t = m \lambda_1 = 3 	imes 420 = 1260 \ nm$.$2 	imes 1.4 	imes t = 1260 \ nm$.$t = 1260 / 2.8 = 450 \ nm$.

$A$ thin film of soap solution $(n = 1.4)$ lies on the top of a glass plate $(n = 1.5)$. When visible light is incident almost normal to the plate,two adjacent reflection maxima are observed at two wavelengths $420 \ nm$ and $630 \ nm$. The minimum thickness of the soap solution is.....$nm$

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