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(C) For maximum transmission of light, the reflected light must undergo destructive interference (minima).Since the refractive index of the film $(\mu

Vedclass · Accepted Answer

$137.5$

Vedclass · Answer

(C) For maximum transmission of light, the reflected light must undergo destructive interference (minima).Since the refractive index of the film $(\mu_f = 2.0)$ is greater than that of air $(\mu_a = 1.0)$ and also greater than that of the glass slab $(\mu_g = 1.45)$, there is a phase change of $\pi$ at both the top and bottom surfaces of the film.The condition for destructive interference (minima) in reflected light is given by $2 \mu_f t = n \lambda$, where $n = 1, 2, 3, \dots$For minimum thickness, we take $n = 1$:$2 \mu_f t = \lambda$$t = \frac{\lambda}{2 \mu_f}$Substituting the given values $\lambda = 550 \ 	ext{nm}$ and $\mu_f = 2.0$:$t = \frac{550 \ 	ext{nm}}{2 	imes 2.0} = \frac{550}{4} \ 	ext{nm} = 137.5 \ 	ext{nm}$.

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