A thin film of water (mu = 4/3) is 3100 ,math | vedclass

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Question

(C) For constructive interference in a thin film (reflected light),the condition is $2 \mu t \cos r = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3,

Vedclass · Accepted Answer

Yellow

Vedclass · Answer

(C) For constructive interference in a thin film (reflected light),the condition is $2 \mu t \cos r = (2n - 1) \frac{\lambda}{2}$,where $n = 1, 2, 3, \dots$ Given $\mu = 4/3$,$t = 3100 \,\mathring{A}$,and normal incidence ($r = 0$,so $\cos r = 1$). The formula becomes $\lambda = \frac{4 \mu t}{2n - 1}$. For $n = 1$: $\lambda = 4 	imes (4/3) 	imes 3100 = 16533 \,\mathring{A}$ (Infrared). For $n = 2$: $\lambda = \frac{4 	imes (4/3) 	imes 3100}{3} = \frac{16533}{3} = 5511 \,\mathring{A}$. This wavelength $(5511 \,\mathring{A})$ corresponds to the yellow region of the visible spectrum. Therefore,the film appears yellow.

$A$ thin film of water $(\mu = 4/3)$ is $3100 \,\mathring{A}$ thick. If it is illuminated by white light at normal incidence,the colour of the film in the reflected light will be:

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