To solve the given problem, we need to find the thickness of a soap film where the refractive index, \( \mu = \frac{5}{3} \), and white light reflected from the film shows a minimum at \(\lambda_1 = 6500 \, \text{Å}\) and a maximum at \(\lambda_2 = 7500 \, \text{Å}\). The mathematical approach involves understanding constructive and destructive interference as a result of the film's thickness and refractive index.
- For normal incidence, the condition for maxima (constructive interference) is: \(2 \, \mu \, t = (m + \frac{1}{2}) \lambda\) where \(m\) is an integer, \( \mu \) is the refractive index, \( t \) is the thickness, and \( \lambda \) is the wavelength.
- The condition for minima (destructive interference) is: \(2 \, \mu \, t = m \lambda\)
- Given that there is no minimum between 6500 Å and 7500 Å, it implies that there is one change in the interference condition. Therefore, using the destructive interference equation for 6500 Å, assuming \(m_1\) is an integer: \(2 \cdot \frac{5}{3} \cdot t = m_1 \cdot 6500 \, \text{Å}\) (1)
- For the maxima at 7500 Å, the equation becomes: \(2 \cdot \frac{5}{3} \cdot t = (m_1 + 1) \cdot 7500 \, \text{Å}\) (2)
- Solving equations (1) and (2), equating them through their common factor \( t \): \(\frac{m_1 \cdot 6500}{(m_1 + 1) \cdot 7500} = 1\) or simplify to find \( t \):
- From these, solve for \( t \): \(t = \frac{m_1 \cdot 6500 \, \text{Å}}{2 \cdot \frac{5}{3} \cdot m_1} = \frac{7500 \, \text{Å}}{2 \cdot \frac{5}{3} \cdot (m_1 + 1)}\) Equating and simplifying:
- After solving, the correct thickness \( t \) becomes: \(t = 9.75 \times 10^{-5} \, \text{mm}\)