In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength l0 in vacuum at normal incidence onto the surface of the glass. When l0=496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

Constructive interference is at $\mathit{n}\mathit{\lambda }$

and destructive interference is at$\frac{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{\lambda }}{\mathbf{2}}$.

Where n is an integer.

A ray will experience a phase change when reflected from a medium of higher refractive index, while no phase change is seen when a ray is reflected against a medium of lower refractive index.

The first reflected ray will experience a phase change since the index of refraction of the glass is greater than the index of refraction of the air. But the second reflected ray will experience NO phase change since the index of refraction of the gas is less than the index of refraction of the glass. So, we have two reflected rays with one phase change, as you see below. Red circle indicates a phase change.

The thickness of the film for constructive interference, for two rays with one phase change, is given by

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

Hence the thickness is given by

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

From snell’s law we have

\({n_1}{\lambda _1} = {n_2}{\lambda _2}\)

And here we get

\({n_0}{\lambda _0} = {n_{film}}{\lambda _{film}}\)

Solve for\({\lambda _{film}}\)

\({\lambda _{film}} = \frac{{{\lambda _0}}}{{{n_{film}}}}\)

Plug this into the thickness formula

\(t = \frac{{(m + \frac{1}{2}){\lambda _0}}}{{2{n_{film}}}}\)

We know that the first constructive interference happens at 386nm and the second one for 496nm.

This means that the constructive interference is for \((m + \frac{1}{2})\) when \({\lambda _0}\)=496nm and the second constructive interference is for \((m + \frac{1}{2}) + 1\) when \({\lambda _0}\)=386nm.

Hence

\(t = \frac{{(m + \frac{1}{2})496}}{{2{n_{film}}}}\)

And

\(t = \frac{{(m + \frac{3}{2})386}}{{2{n_{film}}}}\)

Comparing the two we get

\(\frac{{(m + \frac{1}{2})496}}{{2{n_{film}}}} = \frac{{(m + \frac{3}{2})386}}{{2{n_{film}}}}\)

Solving for m;

\(\begin{array}{l}496m + 28 = 386m + 579\\m = 3\end{array}\)

Plug this into the thickness equation

\(\begin{array}{l}t = \frac{{(3.0 + \frac{1}{2})496}}{{2 \times 1.40}}\\t = 620nm\end{array}\)

Hence the thickness of the film is 620nm

We know

$t=\frac{(m+1/2){\lambda }_0}{2n_{film}}$

Rearranging we get

${\lambda }_0=\frac{2t{n}_{film}}{(m+{\displaystyle \frac{1}{2}})}$

For maximum wavelength plug m=0;

${\lambda }_0=\frac{2t{n}_{fim}}{{\displaystyle \frac{1}{2}}}$

Plug in the values

$$\begin{gathered} {\lambda }_0=4t{n}_{film}=4\times 620\times 1.40 \\ {\lambda }_0=3472nm \end{gathered}$$

Hence,The longest wavelength in vacuum for which there is constructive interference for the reflected light is 3472nm