Question

A thin film suspended in air is 0.410 $\mathbf{\mu m}$thick and is illuminated with white light incident perpendicularly on its surface. The index of refraction of the film is 1.50. At what wavelength will visible light that is reflected from the two surfaces of the film undergo fully constructive interference?

Short answer

Thus, the wavelength of visible light is $492nm$.

Step-by-step solution

According to the question.

The formula for the constructive interference:

$\mathbf{2}{\mathbf{n}}_{\mathbf{2}}\mathbf{L}\mathbf{=}\left(m+\frac{1}{2}\right)\mathbf{\lambda }$

Here,$\mathbf{\lambda }$is wavelength of light in air/vacuum.

The wavelength of visible light. 

Use the above formula as follows:

$$\begin{gathered} \lambda =\frac{2n_{2}L}{m+\frac{1}{2}} \\ =\frac{2\left(1.50\right)\left(410 \text{nm}\right)}{m+\frac{1}{2}} \\ =\frac{1230\text{nm}}{m+\frac{1}{2}} \end{gathered}$$

Where $m=0,1,2,...$The only value of $m$which, when substituted into equation above, would yield a wavelength that within the visible light range is$m=1$ .

Therefore, the wavelength is:

$$\begin{gathered} \lambda =\frac{1230 \text{nm}}{1+\frac{1}{2}} \\ =492 \text{nm} \end{gathered}$$

Hence, the wavelength of visible light is$492nm$.