Light of wavelength 6000 , mathring{A} is inc | vedclass

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(C) For a thin film to appear dark in reflection,the condition for destructive interference is given by $2 \mu t \cos r = n \lambda$,where $n = 1, 2,

Vedclass · Accepted Answer

$4 	imes 10^{-7} \, m$

Vedclass · Answer

(C) For a thin film to appear dark in reflection,the condition for destructive interference is given by $2 \mu t \cos r = n \lambda$,where $n = 1, 2, 3, ...$ Taking the minimum thickness $(n = 1)$,the formula becomes $t = \frac{n \lambda}{2 \mu \cos r}$. Given: $\lambda = 6000 \, \mathring{A} = 6 	imes 10^{-7} \, m$,$\mu = 1.5$,and $r = 60^\circ$. Substituting the values: $t = \frac{1 	imes 6 	imes 10^{-7}}{2 	imes 1.5 	imes \cos 60^\circ}$. Since $\cos 60^\circ = 0.5$,we have $t = \frac{6 	imes 10^{-7}}{3 	imes 0.5} = \frac{6 	imes 10^{-7}}{1.5} = 4 	imes 10^{-7} \, m$.

Light of wavelength $6000 \, \mathring{A}$ is incident on a thin glass plate of refractive index $1.5$ such that the angle of refraction in the plate is $60^\circ$. Find the thickness of the plate so that it appears dark in reflection.

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