Question

A soap film has an index of refraction \(n=1.50 .\) The film is viewed in reflected light. (a) At a spot where the film thickness is $910.0 \mathrm{nm},$ which wavelengths are missing in the reflected light? (b) Which wavelengths are strongest in reflected light?

Short answer

Answer: The missing wavelengths in the reflected light are approximately 910 nm, 455 nm, and 303.3 nm. The wavelength with the strongest reflection in the visible range is approximately 364 nm.

Step-by-step solution

(a) Find the wavelengths missing in the reflected light (destructive interference)

To find the missing wavelengths in the reflected light, we will consider destructive interference. The condition for destructive interference is given by: \(m\lambda = 2t(n - 1)\) where \(m\) is an integer (the order of interference). We are given the thickness of the film, \(t = 910 \ \mathrm{nm}\), and the index of refraction, \(n = 1.50\). The goal is to find the wavelengths \(\lambda\) that satisfy this condition for given values of \(m\). First, let's rearrange the interference condition formula to solve for \(\lambda\): \(\lambda = \frac{2t(n - 1)}{m}\) Now, let's plug in the given values and find the wavelengths for different values of \(m\). We will start with \(m = 1, 2, 3,...\) and keep going until \(\lambda\) becomes smaller than the visible range of wavelengths (about 400 nm to 700 nm). For \(m=1\): \(\lambda = \frac{2(910 \ \mathrm{nm})(1.50 - 1)}{1} = \text{910 nm}\) For \(m=2\): \(\lambda = \frac{2(910 \ \mathrm{nm})(1.50 - 1)}{2} = \text{455 nm}\) For \(m=3\): \(\lambda = \frac{2(910 \ \mathrm{nm})(1.50 - 1)}{3} = \text{303\overline{3} nm}\) Thus, the missing wavelengths in the reflected light are approximately \(\text{910 nm, 455 nm}\) and \(\text{303}\overline{3} \ \mathrm{nm}\).

(b) Find the wavelengths strongest in reflected light (constructive interference)

To find the strongest wavelengths in the reflected light, we will consider constructive interference. The condition for constructive interference is given by: \((m + \frac{1}{2})\lambda = 2t(n - 1)\) Again, we are given the thickness of the film, \(t = 910 \ \mathrm{nm}\), and the index of refraction, \(n = 1.50\). The goal is to find the wavelengths \(\lambda\) that satisfy this condition for given values of \(m\). First, let's rearrange the interference condition formula and solve for \(\lambda\): \(\lambda = \frac{2t(n - 1)}{m + \frac{1}{2}}\) Now, let's plug in the given values and find the wavelengths for different values of \(m\). We will start with \(m = 1, 2, 3,...\) and continue until \(\lambda\) becomes smaller than the visible range of wavelengths (about 400 nm to 700 nm). For \(m=1\): \(\lambda = \frac{2(910 \ \mathrm{nm})(1.50 - 1)}{1 + \frac{1}{2}} = \text{121\overline{3} nm}\) For \(m=2\): \(\lambda = \frac{2(910 \ \mathrm{nm})(1.50 - 1)}{2 + \frac{1}{2}} = \text{364 nm}\) For \(m=3\): \(\lambda = \frac{2(910 \ \mathrm{nm})(1.50 - 1)}{3 + \frac{1}{2}} \approx \text{242.\overline{8} nm}\) Since \(\text{121}\overline{3} \ \mathrm{nm}\) and \(\text{242.\overline{8}} \ \mathrm{nm}\) are out of the visible range, the wavelength with the strongest reflection in the visible range is approximately \(\text{364 nm}\).