Let's break down the steps to solve the problem. --- ### **Step 1: Write the Reflection Coefficient for p-polarized Light (\( R_p \))** For p-polarized light, the reflection coefficient at the boundary between two media with refractive indices \( n_1 \) (medium, dielectric constant \( K = n_1^2 \)) and \( n_2 \) (air, \( n_2 \approx 1 \)) is: \[ R_p = \left| \frac{n_2 \cos \theta_1 - n_1 \cos \theta_2}{n_2 \cos \theta_1 + n_1 \cos \theta_2} \right|^2 \] --- ### **Step 2: Use Snell's Law** \[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \] At the **critical angle** (\( \theta_c \)), \( \theta_2 = 90^\circ \): \[ n_1 \sin \theta_c = n_2 \sin 90^\circ \implies \sin \theta_c = \frac{n_2}{n_1} \] Let \( \theta_1 = \theta_c - \delta \) (with \( \delta \to 0^+ \)). --- ### **Step 3: Approximate \(\sin \theta_1\) and \(\cos \theta_1\)** For small \( \delta \): - \( \sin \theta_1 \approx \sin \theta_c - \delta \cos \theta_c \) - \( \cos \theta_1 \approx \cos \theta_c + \delta \sin \theta_c \) --- ### **Step 4: Find \( \cos \theta_2 \) for \( \theta_1 < \theta_c \) (Just Below Critical Angle)** From Snell's Law: \[ \sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1 \] Since \( \theta_1 < \theta_c \), \( \sin \theta_1 < \sin \theta_c \), so \( \sin \theta_2 < 1 \) and \( \theta_2 \) is real. \[ \cos \theta_2 = \sqrt{1 - \sin^2 \theta_2} \] --- ### **Step 5: Expand \( \sin \theta_1 \) and \( \sin \theta_2 \) to First Order in \( \delta \)** \[ \sin \theta_1 \approx \sin \theta_c - \delta \cos \theta_c \] \[ \sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1 = \frac{n_1}{n_2} (\sin \theta_c - \delta \cos \theta_c) \] But \( \sin \theta_c = \frac{n_2}{n_1} \), thus: \[ \sin \theta_2 = 1 - \frac{n_1}{n_2} \delta \cos \theta_c \] So, \[ 1 - \sin^2 \theta_2 \approx 1 - [1 - 2\frac{n_1}{n_2} \delta \cos \theta_c] = 2 \frac{n_1}{n_2} \delta \cos \theta_c \] Thus, \[ \cos \theta_2 \approx \sqrt{2 \frac{n_1}{n_2} \delta \cos \theta_c} \] --- ### **Step 6: Substitute These Approximations into \( R_p \)** Recall: \[ R_p = \left| \frac{n_2 \cos \theta_1 - n_1 \cos \theta_2}{n_2 \cos \theta_1 + n_1 \cos \theta_2} \right|^2 \] Let’s use the approximations: - \( n_2 \cos \theta_1 \approx n_2 \cos \theta_c \) - \( n_1 \cos \theta_2 \approx n_1 \sqrt{2 \frac{n_1}{n_2} \delta \cos \theta_c} = \sqrt{2 n_1^3 \delta \cos \theta_c / n_2} \) So, \[ R_p \approx \left| \frac{n_2 \cos \theta_c - \sqrt{2 n_1^3 \delta \cos \theta_c / n_2}}{n_2 \cos \theta_c + \sqrt{2 n_1^3 \delta \cos \theta_c / n_2}} \right|^2 \] --- ### **Step 7: Analyze the Slope at \( \theta_1 = \theta_c \) (\( \delta = 0 \))** As \( \delta \to 0 \): - \( n_1 \cos \theta_2 \to 0 \) - \( R_p \to 1 \) Now, compute the derivative \( \frac{dR_p}{d\theta_1} \) as \( \delta \to 0 \): Let’s write \( R_p \approx \left(1 - A \sqrt{\delta} \right)^2 \) for small \( \delta \), where \( A = \frac{\sqrt{2 n_1^3 \cos \theta_c / n_2}}{n_2 \cos \theta_c} \): \[ R_p \approx \left(1 - A \sqrt{\delta}\right)^2 = 1 - 2A \sqrt{\delta} + A^2 \delta \] So, \[ \frac{dR_p}{d\delta} \approx -A \delta^{-1/2} \] As \( \delta \to 0 \), \( \frac{dR_p}{d\delta} \to -\infty \). --- ### **Step 8: Final Answers** #### **a) \( R_p \) as a Function of \( \delta \) as \( \delta \to 0 \):** \[ \boxed{ R_p \approx 1 - 2A \sqrt{\delta} } \] where \[ A = \frac{\sqrt{2 n_1^3 \cos \theta_c / n_2}}{n_2 \cos \theta_c} \] #### **b) Slope is Infinite at \( \theta_c \):** The slope \( \frac{dR_p}{d\theta_1} \) diverges as \( \delta \to 0 \): \[ \boxed{ \text{The slope of the } R_p(\theta_1) \text{ curve is infinite at } \theta_1 = \theta_c. } \] --- **Summary Table:** | Step | Result | |--------------------------------------|---------------------------------------------------| | \( R_p \) as \( \delta \to 0 \) | \( R_p \approx 1 - 2A \sqrt{\delta} \) | | Slope at \( \theta_1 = \theta_c \) | Infinite (\( \frac{dR_p}{d\theta_1} \to \infty \))| If you'd like, I can substitute explicit values for \( n_1 \), \( n_2 \), and \( \theta_c \) for a specific case.