White light is sent downward onto a horizontal thin film that is sandwiched between two materials. The indexes of refraction are $\mathbf{1}.\mathbf{80}$for the top material, $\mathbf{1}.\mathbf{70}$for the thin film, and $\mathbf{1}.\mathbf{50}$for the bottom material. The film thickness is$\mathbf{5}\times {\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{m}$ . Of the visible wavelengths ($\mathbf{400}$ to $\mathbf{700}\mathbf{}\mathbf{nm}$ ) that result in fully constructive interference at an observer above the film, which is the (a) longer and (b) shorter wavelength? The materials and film are then heated so that the film thickness increases. (c) Does the light resulting in fully constructive interference shift toward longer or shorter wavelengths?

The index of refraction for top material is $n_t=1.80$

The index of refraction for bottom material is $n_b=1.50$.

The index of refraction for thin film is $n=1.70$.

The thickness of thin film is $t=5\times {10}^{-7}\text{m}$.

The thickness of the thin film in interference varies with the index of refraction, wavelength of light and order of fringe.

The wavelength of the light for constructive interference is given as:

${\lambda }_m=\frac{2nt}{m}$ (i)

Substitute $m=1$in the above equation (i).

$\begin{array}{rcl}{\lambda }_1& =& \frac{2\left(1.70\right)\left(5\times {10}^{-7}\text{m}\right)}{1}\\ & =& 17\times {10}^{-7}\text{m}\\ & =& \left(17\times {10}^{-7}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ & =& 1700\text{nm}\end{array}$

Substitute $m=2$in the above equation (ii).

$\begin{array}{rcl}{\lambda }_2& =& \frac{2\left(1.70\right)\left(5\times {10}^{-7}\text{m}\right)}{2}\\ & =& 8.50\times {10}^{-7}\text{m}\\ & =& \left(8.50\times {10}^{-7}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ & =& 850\text{nm}\end{array}$

Substitute $m=3$in the above equation (i).

$\begin{array}{rcl}{\lambda }_3& =& \frac{2\left(1.70\right)\left(5\times {10}^{-7}\text{m}\right)}{3}\\ & =& 5.67\times {10}^{-7}\text{m}\\ & =& \left(5.67\times {10}^{-7}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ & =& 567\text{nm}\end{array}$

The wavelengths for first and second order fringes are more than the maximum value of wavelength of visible range of light. The wavelength for third order fringe is longer wavelength for constructive interference in visible range because it is next longer wavelength inside visible range of light.

Therefore, the longer wavelength for constructive interference is $567\text{nm}$.

Substitute $m=4$in the above equation (i).

$\begin{array}{rcl}{\lambda }_4& =& \frac{2\left(1.70\right)\left(5\times {10}^{-7}\text{m}\right)}{4}\\ & =& 4.25\times {10}^{-7}\text{m}\\ & =& \left(4.25\times {10}^{-7}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ & =& 425\text{nm}\end{array}$

Substitute $m=5$in the above equation (i).

$\begin{array}{rcl}{\lambda }_5& =& \frac{2\left(1.70\right)\left(5\times {10}^{-7}\text{m}\right)}{5}\\ & =& 3.40\times {10}^{-7}\text{m}\\ & =& \left(3.40\times {10}^{-7}\text{m}\right)\left(\frac{1\text{nm}}{{10}^{-9}\text{m}}\right)\\ & =& 340\text{nm}\end{array}$

The wavelength for fifth order fringe is less than the minimum value of wavelength of visible range of light. The wavelength for fourth order fringe is shorter wavelength for constructive interference in visible range because it is next shorter wavelength inside visible range of light.

Therefore, the shorter wavelength for constructive interference is $425\text{nm}$.

The index of refraction of thin film remains unaffected by the variation in temperature so the resulting light in constructive interference shifts towards longer wavelength.

Therefore, the resulting light in constructive interference shifts towards longer wavelength.