91Ó°ÊÓ

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 \(\text{303}\overline{3} \ \mathrm{nm}\).

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 \(\text{364 nm}\).