In a setup similar to that of Problem 35.39, the glass has an index of refraction of 1.53, the plates are each 8.00 cm long, and the metal foil is 0.015 mm thick. The space between the plates is filled with a jelly whose refractive index is not known precisely but is known to be greater than that of the glass. When you illuminate these plates from above with light of wavelength 525 nm, you observe a series of equally spaced dark fringes in the reflected light. You measure the spacing of these fringes and find that there are 10 of them every 6.33 mm. What is the index of refraction of the jelly?

$$\begin{gathered} n_{plates}=1.53 \\ L=8.0cm=8.0\times {10}^{-2}m \\ {t}_{foil}=0.015mm=0.015\times {10}^{-3}m \\ {n}_{jelly}>n_{plates} \\ {\lambda }_{air}=525nm=525\times {10}^{-9}m \\ {N}_{darkfringes}=10darkfringe \\ x=6.33mm=6.33\times {10}^{-3}m \end{gathered}$$

We know that the index of refraction of the jelly is greater than the index of refraction of the plates. This means that the first reflected ray, as you see in the figure above, will experience NO phase change but the second reflected ray will experience a phase change. (The red circle indicates a phase change)

We need to find the thicknesses in which the two reflected rays of the 525 nm- wavelength will interact destructively.

Since we have one phase change, so the thickness is given by

\[2t = m{\lambda _{film}}\]

For this case,

\[2t = m{\lambda _{jelly}}\] (1)

We know, from Snell's law, that

\[{n_1}{\lambda _1} = {n_2}{\lambda _2}\]

For this case,

\[{n_{air}}{\lambda _{air}} = {n_{jelly}}{\lambda _{jelly}}\]

Solving for\[{\lambda _{jelly}}\]and noting that\[{n_{air}}\]= 1.0

\[{\lambda _{jelly}} = \frac{{{\lambda _{air}}}}{{{n_{jelly}}}}\]

Substitute into (1)

\[2t = \frac{{m{\lambda _{air}}}}{{{n_{jelly}}}}\] (2)

Equation (2) above indicates a zero thickness when m= 0 for the first dark fringe. This is the point where the two plates are in contact, as you see on the left side of the figure above.

Now we need to find the thickness t in terms of x and\[\theta \], whereas t is the distance between the contact point and any dark fringe and is the angle between the two plates.

From the right triangle in the figure below

\[\tan \theta = \frac{t}{x}\]

Hence,

\[t = x\tan \theta \]

Substitute into (2)

\[2x\tan \theta = \frac{{m{\lambda _{air}}}}{{{n_{jelly}}}}\]

We can easily find\[\tan \theta \]by using the larger triangle, whereas x= L and t =\[{t_{foil}}\]

So,

\[\tan \theta = \frac{{{t_{foil}}}}{L}\] (4)

Solving (3) for\[{n_{jelly}}\]

\[{n_{jelly}} = \frac{{m{\lambda _{air}}}}{{2x\tan \theta }}\]

Substitute \[\tan \theta \] form (4);

\[{n_{jelly}} = \frac{{m{\lambda _{air}}L}}{{2x{t_{foil}}}}\]

We know, from (2) above, that the first dark fringe is at the contact point (from the left side in the figure above) which for m = 0.

And we also know that the distance of 6.33 mm contains 10 dark fringes, so the tenth dark fringe is at a distance of 633 mm from the contact point

Hence,$x=6.33\times {10}^{-3}mm$ when m = 10.

Substituteinto (4) andsubstitutethe given;

$$\begin{gathered} n_{jelly}=\frac{10\times 25\times {10}^{-9}\times 8.0\times {10}^{-2}}{2\times 6.33\times {10}^{-3}\times 0.015\times {10}^{-3}} \\ {n}_{jelly}=2.21 \end{gathered}$$