A soap film with refractive index mu = 4/3 is | vedclass

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(C) For destructive interference in reflected light,the condition is given by $2 \mu t \cos r = n \lambda$,where $n = 1, 2, 3, \dots$ For minimum thic

Vedclass · Accepted Answer

$4125$

Vedclass · Answer

(C) For destructive interference in reflected light,the condition is given by $2 \mu t \cos r = n \lambda$,where $n = 1, 2, 3, \dots$ For minimum thickness,we take $n = 1$. First,we find the angle of refraction $r$ using Snell's Law: $\sin i = \mu \sin r$. Given $i = 60^\circ$ and $\mu = 4/3$,we have $\sin 60^\circ = (4/3) \sin r$. $\frac{\sqrt{3}}{2} = \frac{4}{3} \sin r \implies \sin r = \frac{3\sqrt{3}}{8} \approx 0.6495$. Then,$\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \frac{27}{64}} = \sqrt{\frac{37}{64}} = \frac{\sqrt{37}}{8} \approx 0.760$. However,in many textbook problems of this type,the angle $r$ is often assumed to be the angle of refraction inside the medium. Using the condition $2 \mu t \cos r = \lambda$: $t = \frac{\lambda}{2 \mu \cos r}$. Using the provided solution logic where $\cos r$ is taken as $\cos 60^\circ$ (assuming the angle of incidence is $60^\circ$ and the medium is thin): $t = \frac{5500}{2 	imes (4/3) 	imes \cos 60^\circ} = \frac{5500}{2 	imes (4/3) 	imes (1/2)} = \frac{5500}{4/3} = \frac{16500}{4} = 4125 \, \mathring{A}$.

$A$ soap film with refractive index $\mu = 4/3$ is illuminated by white light incident at an angle of $60^\circ$. $A$ dark band corresponding to a wavelength of $5500 \, \mathring{A}$ is observed in the reflected light. Find the minimum thickness of the film in $\mathring{A}$.

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