A uniform film of TiO2 , 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 520.0 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels.

(a) What is the minimumthickness of TiO2 that you must addso the reflected light cancels as desired?

(b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wavelengths of the light in the TiO2 film.

As you see in the figure above, the first ray experiences a phase change since the index of refraction of air is less than the index of refraction of the film. But the second ray does not experience a phase change since the index of refraction of the film is greater than that of the crown glass.

Hence, we have two reflected rays with one phase change.

The red circle, in the figure above, indicates a phase change.

So to find the thickness of the film that causes a destructive inference, we need to use the following formula. $$\begin{gathered} 2\mathrm{t}={\mathrm{m\lambda }}_{\mathrm{film}} \\ \mathrm{t}=\frac{{\mathrm{m\lambda }}_{\mathrm{film}}}{2} \end{gathered}$$

We know, from Snell's law, that

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

So,

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

So for the case,

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

And we also know that

role="math" localid="1664002089842" ${\lambda }_{film}=\frac{n_{air}}{n_{film}}$ (A)

Substitute into (1)

$t=\frac{m\lambda air}{2n_{film}}$

Substitute the given,

$$\begin{gathered} \mathrm{t}=\frac{520.0\times {10}^{-9}\mathrm{m}}{2\times 2.62} \\ \mathrm{t}=9.92\times {10}^{-8}\mathrm{m} \\ \mathrm{t}=99.2\mathrm{m\_nm} \end{gathered}$$ (2)

Noting that the thickness found in equation (2) is the whole thickness needed to produce a destructive interference

We need to find the extra thickness we need to add to the initial thickness, but first, we need to find the m for this new thickness.

If we used m = 10, then the final thickness will be

$t_{\left(m=10\right)}=992nm$

which is still less than the initial thickness coated above the glass.

So, we will use the lth dark fringe. Hence,

$$\begin{gathered} t=99.2\times 11 \\ t=1091nm \end{gathered}$$ (3)

Noting that the extra thickness is given by

$$\begin{gathered} t=t_i+t_{extra} \\ {t}_{extra}=t-t_i \end{gathered}$$

Substitute from (2) and from the given above.

$$\begin{gathered} t_{extra}=\left(1091\times {10}^{-9}\right)-\left(1036\times {10}^{-9}\right) \\ {t}_{extra}=55.0nm \end{gathered}$$

The path difference is given by the double of the total thickness of the film.

Hence,

$△\mathrm{r}=2\mathrm{t}△\mathrm{r}=2\mathrm{t}$

Substitute from (3)

$$\begin{gathered} △r=2\times 1091nm△r=2182nm \\ △r=2182nm \end{gathered}$$

The wavelength of the light in the film is given by equation (A) above.

so the path difference in terms of the wavelength of the light in the film is given by,

$△r=\frac{2t}{{\lambda }_{film}}\times {\lambda }_{film}$

Substitute the ${\lambda }_{film}$in the denominator from (A).

Hence,

$△r=\frac{2t{n}_{film}}{{\lambda }_{air}}\times {\lambda }_{film}$

Substitute the given and (2)

$$\begin{gathered} △r=\frac{2\times 1091\times 2.62}{520}\times {\lambda }_{film} \\ △r=11.0{\lambda }_{film} \end{gathered}$$