91Ó°ÊÓ

$n_{benzene}=n_{film}=1.50,{\mathrm{λ}}_{air}=575nm$

The first reflected ray, as you see in the figure above, experience a phase change since the index f refraction of the oil is greater than the index of refraction of the air.

(The red circle indicates a phase change)

But the second ray reflects without any phase change since the index of refraction water is less than the index of refraction of the oil

This means that we have two reflected rays with one phase change.

So, the thickness of the film, which gives a constructive interference at the wavelength of 575 nm, is given by

$2t=\left(m+\frac{1}{2}\right){\mathrm{λ}}_{film}$ (1)

We know, from Snell's law, that

$n_1{\mathrm{λ}}_1=n_2{\mathrm{λ}}_2$

So,

$n_{film}{\mathrm{λ}}_{film}=n_{air}{\mathrm{λ}}_{air}$

whereas$n_{air}$= 1.0 Solving for${\mathrm{λ}}_{film}$and

${\mathrm{λ}}_{film}=\frac{n_{air}}{n_{film}}$

Substitute into (1)

$2t=\left(m+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{air}}{{\mathrm{λ}}_{film}}$

Finding the minimum thickness, which occurs when m = 0,

$$\begin{gathered} 2t=\left(0+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{air}}{n_{film}} \\ 2t=\frac{{\mathrm{λ}}_{air}}{2n_{film}} \end{gathered}$$

Hence,

$t=\frac{{\mathrm{λ}}_{air}}{4n_{film}}$

Substitute the given

$$\begin{gathered} t=\frac{575nm}{4×1.50} \\ t=95.8nm \end{gathered}$$