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(A) For a thin film, the condition for destructive interference (minima) in transmitted light is given by $2 \mu t = n \lambda$, where $n = 1, 2, 3, \

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(A) For a thin film, the condition for destructive interference (minima) in transmitted light is given by $2 \mu t = n \lambda$, where $n = 1, 2, 3, \dots$.Here, $\mu = 1.4$ and $\lambda = 560 	imes 10^{-9} \ m$.The change in thickness $\Delta t$ for consecutive minima is $\Delta t = \frac{\lambda}{2 \mu}$.Substituting the values: $\Delta t = \frac{560 	imes 10^{-9}}{2 	imes 1.4} = \frac{560 	imes 10^{-9}}{2.8} = 200 	imes 10^{-9} \ m = 2 	imes 10^{-7} \ m$.The area of the film is $A = \pi r^2 = \pi (1.8 	imes 10^{-2} \ m)^2 = \pi 	imes 3.24 	imes 10^{-4} \ m^2$.The rate of evaporation is $R = \frac{A 	imes \Delta t}{\Delta T}$, where $\Delta T = 12 \ s$.$R = \frac{\pi 	imes 3.24 	imes 10^{-4} 	imes 2 	imes 10^{-7}}{12} = \frac{\pi 	imes 6.48 	imes 10^{-11}}{12} = 0.54 	imes 10^{-11} \ m^3/s = 54 	imes 10^{-13} \ m^3/s$.Thus, the rate of evaporation is $54 \pi 	imes 10^{-13} \ m^3/s$.

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