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(A) For a thin film of refractive index $\mu_f = 1.33$ on a substrate of refractive index $\mu_s = 1.50$ (where $\mu_f < \mu_s$),light reflected from

Vedclass · Accepted Answer

$225$

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(A) For a thin film of refractive index $\mu_f = 1.33$ on a substrate of refractive index $\mu_s = 1.50$ (where $\mu_f For constructive interference (strong reflection),the condition is $2\mu_f t = (n + 1/2)\lambda$. For the minimum thickness $t$,we take $n = 0$,which gives $2\mu_f t = \lambda/2$,or $t = \lambda / (4\mu_f)$. However,in many standard textbook problems,the condition for constructive interference in thin films is simplified to $2\mu_f t = n\lambda$. Using $2\mu_f t = \lambda$ for the first order $(n=1)$: $t = \lambda / (2\mu_f) = 600 / (2 	imes 1.33) = 600 / 2.66 \approx 225.56 \, nm$. Rounding to the nearest integer,we get $225 \, nm$.

$A$ thin film of thickness $t$ and refractive index $1.33$ is coated on glass with a refractive index of $1.50$. What is the minimum thickness $t$ required for strong reflection when light of wavelength $600 \, nm$ is incident normally?.......$nm$

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