Let's solve both parts step-by-step: --- ## **a. Polarizing Sheets** ### **i. Orientations of Polarizers 1 and 3** Given: - There are three polarizing sheets. - The middle sheet (#2) can be rotated, and its angle is denoted by θ₂. - When θ₂ = 90°, the plot shows maximum intensity, so polarizers 1 and 3 must be parallel. **Orientations:** - Let’s set the y-axis as 0°. - Given the plot, **polarizer 1** is at 0°. - When θ₂ = 90°, maximum intensity means **polarizer 3** is also at 90°. - But generally: - **Polarizer 1:** 0° - **Polarizer 2:** θ₂ - **Polarizer 3:** 90° --- ### **ii. Sketch Situation at θ₂ = 20°** **Drawing:** - Draw three sheets. - Sheet 1: vertical lines (0°) - Sheet 2: lines at 20° from vertical - Sheet 3: horizontal lines (90°) - Light polarization: - After Sheet 1: vertical (0°) - After Sheet 2: 20° - After Sheet 3: 90° (horizontal) --- ### **iii. Relative Intensity After All Three Sheets** **Formula:** - Unpolarized light through first polarizer: \( I_1 = \frac{I_0}{2} \) - After sheet 2 (Malus' Law): \( I_2 = I_1 \cos^2(\theta_2 - \theta_1) = \frac{I_0}{2} \cos^2(20°-0°) \) - After sheet 3: \( I_3 = I_2 \cos^2(\theta_3 - \theta_2) = \frac{I_0}{2} \cos^2(20°) \cos^2(90°-20°) \) - \( \cos^2(70°) \approx 0.116 \), \( \cos^2(20°) \approx 0.883 \) **Calculation:** \[ I_3 = \frac{I_0}{2} (0.883)(0.116) \approx \frac{I_0}{2} \times 0.102 \approx 0.051 I_0 \] **Relative Intensity \( I_{rel} = \frac{I_3}{I_0} \):** \[ I_{rel} \approx 0.051 \] --- ## **b. Refraction and Critical Angle** ### **i. Find Critical Angle and Angle of Refraction** #### **Critical Angle at Interface (n₂ to n₃):** - \( n_2 = 1.80 \), \( n_3 = 1.30 \) - Critical angle \( \theta_c \) is defined by: \[ \sin \theta_c = \frac{n_3}{n_2} = \frac{1.30}{1.80} = 0.722 \] \[ \theta_c = \arcsin(0.722) \approx 46.1° \] So, **θ = 46.1°**. #### **Angle of Refraction into Material 4 (n₄ = 1.40):** - At critical angle, light *just grazes* the interface, so it refracts into n₄. - Use Snell's Law: \[ n_2 \sin \theta = n_4 \sin \theta_4 \] \[ 1.80 \sin(46.1°) = 1.40 \sin \theta_4 \] \[ 1.80 \times 0.722 = 1.40 \sin \theta_4 \] \[ 1.2996 = 1.40 \sin \theta_4 \] \[ \sin \theta_4 = \frac{1.2996}{1.40} \approx 0.928 \] \[ \theta_4 = \arcsin(0.928) \approx 68.3° \] --- ### **ii. Accurate Diagram** **Draw four regions:** - 1 (n₁ = 1.60): leftmost region, light enters at angle θ. - 2 (n₂ = 1.80): thin layer, same angle θ. - 3 (n₃ = 1.30): above layer 2, light arrives at the critical angle. - 4 (n₄ = 1.40): right region, light refracts at θ₄. Show ray going from 1 → 2 → at the interface between 2 and 3 at θ = 46.1°, and then refracting into 4 at θ₄ = 68.3°. --- ### **iii. If θ is Decreased** - If **θ is decreased below the critical angle**, total internal reflection does **not** occur. - Therefore, **some light refracts into material 3** (n₃), not just into material 4. --- ## **Final Answers** ### **a.** (i) **Orientations:** - Polarizer 1: 0°, Polarizer 2: θ₂, Polarizer 3: 90° (ii) **Sketch:** - Show three sheets at 0°, 20°, 90°; polarization direction changes at each sheet. (iii) **Relative Intensity:** \[ I_{rel} \approx 0.051 \] --- ### **b.** (i) **Critical angle θ:** \( 46.1° \); **Angle of refraction into n₄:** \( 68.3° \) (ii) **Diagram:** - Show four regions, ray path as described. (iii) **If θ is decreased:** - **Yes,** there is refraction into material 3. --- **Let me know if you'd like any step explained further or want a hand-drawn sketch!**