White light reflects at normal incidence from the top and bottom surfaces of a glass plate 1n= 1.522. There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 nm. What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 nm?

The first reflected ray will experience a phase change since the index of refraction of the air is less than the index of refraction of the glass, as you see in the figure below.

The red circle indicates a phase change.

The second reflected ray will reflect without any phase change, for the same reason above and since it travels through the glass.

From all of that, we have two reflected rays with one phase change.

So, the thickness of the film for constructive interference for this case (one phase change) is given by

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

Apply snell's law in the given situation

We know, from Snell's law, that

$n_1{\lambda }_1=n_2{\lambda }_2$

For our case,

$n_{air}{\lambda }_{air}=n_{film}{\lambda }_{film}$

Solving for${\lambda }_{film}$and remembering that=1

${\lambda }_{film}=\frac{{\lambda }_{air}}{n_{film}}$

Substitute into (1)

$2t=\left(m+\frac{1}{2}\right)\times \frac{{\lambda }_{air}}{n_{film}}$

Hence,

$t=\left(m+\frac{1}{2}\right)\times \frac{{\lambda }_{air}}{2n_{film}}$

Now we will use the last equation (2), for both cases. Noting that, the first bright fringe ofoccurs at mi, and the first bright fringe ofoccurs at

Hence,

$$\begin{gathered} t=\left(m_1+\frac{1}{2}\right)\times \frac{{\lambda }_{air,1}}{2n_{film}} \\ t=\left(m_1+\frac{1}{2}\right)\times \frac{{\lambda }_{air,2}}{2n_{film}} \end{gathered}$$

….eq. (3) and (4) respectively.

These last two equations, (3) and (4) are equal since the thickness of the glass film is stable.

Hence,

$2n_{film}$cancels, $\left(m_1+\frac{1}{2}\right)\times \frac{{\lambda }_{air,1}}{2n_{film}}=\left(m_2+\frac{1}{2}\right)\times \frac{{\lambda }_{air,2}}{2nfilm}$

$\left(m_1+\frac{1}{2}\right)\times \frac{{\lambda }_{air,1}}{1}=\left(m_2+\frac{1}{2}\right)\times \frac{{\lambda }_{air,2}}{1}$ …(5)

We know that the two first bright fringes of the two wavelengths are next to each other(here are two adjacent bright fringes).

This means that the first bright fringe of the second wavelengthis the second bright fringe of the system.

Hence,

$$\begin{gathered} m_1-m_2=1 \\ {m}_2=m_1-1 \end{gathered}$$

Substitute into (5)

$$\begin{gathered} \left(m_1+\frac{1}{2}\right)\times {\lambda }_{air,1}=\left(m_1-1+\frac{1}{2}\right)\times {\lambda }_{air,2} \\ \left(m_1+\frac{1}{2}\right)\times {\lambda }_{air,1}=\left(m_1-\frac{1}{2}\right)\times {\lambda }_{air,2} \end{gathered}$$

Solving for

$$\begin{gathered} m_1{\lambda }_{air,1}+\frac{{\lambda }_{air,1}}{2}=m_1{\lambda }_{air,2}-\frac{{\lambda }_{air,2}}{2} \\ {m}_1{\lambda }_{air,1}-m_1{\lambda }_{air,2}=-\frac{{\lambda }_{air,2}}{2}-\frac{{\lambda }_{air,1}}{2} \\ {m}_1\left[{\lambda }_{air,1}-{\lambda }_{air,2}\right]=-\left(\frac{{\lambda }_{air,2}+{\lambda }_{air,1}}{2}\right) \\ {m}_1=-\left(\frac{{\lambda }_{air,2}+{\lambda }_{air,1}}{2\left[{\lambda }_{air,1}-{\lambda }_{air,2}\right]}\right) \end{gathered}$$

Substitute the given above,

$$\begin{gathered} m_1=-\left(\frac{540.6+477}{2\left[477-540.6\right]}\right) \\ {m}_1=8 \end{gathered}$$

Substitute into (3),

$t=\left(8+\frac{1}{2}\right)\times \frac{{\lambda }_{air,1}}{2n_{film}}$

Substitute the given,

$$\begin{gathered} t=\left(8+\frac{1}{2}\right)\times \frac{477.0}{2\times 1.52} \\ t=1334nm \end{gathered}$$