Let's solve each part step by step, with explanations for each step and a final summary. --- ### **Given Data and Definitions** - Thickness of glass block, \( t = 2.00 \) cm \( = 0.0200 \) m - Index of refraction of glass, \( n = 1.98 \) - Angle of incidence, \( \theta_1 = 39.6^\circ \) - Index of refraction of air, \( n_1 \approx 1.00 \) - Speed of light in vacuum, \( c = 3.00 \times 10^8 \) m/s --- ## **Step 1: Find the angle of refraction at the top surface (\( \theta_2 \)) and the angle of incidence at the bottom surface** ### **Calculation** **Snell's Law:** \[ n_1 \sin\theta_1 = n_2 \sin\theta_2 \] Substitute values: \[ 1.00 \times \sin(39.6^\circ) = 1.98 \times \sin\theta_2 \] \[ \sin\theta_2 = \frac{\sin(39.6^\circ)}{1.98} \] \[ \sin\theta_2 = \frac{0.6376}{1.98} = 0.322 \] \[ \theta_2 = \arcsin(0.322) = 18.8^\circ \] **Explanation:** Snell's Law relates the angles and refractive indices of the two media. The angle of incidence at the top surface is given, and the angle of refraction inside the glass (\( \theta_2 \)) is calculated. --- ## **Step 2: Find the refracted angle at the bottom surface** ### **Calculation** The angle at which the light hits the bottom surface inside the glass is the same as the angle of refraction found above, so: \[ \theta_2 = 18.8^\circ \] When the light exits the glass block into air, apply Snell's Law again: \[ n_2 \sin\theta_2 = n_1 \sin\theta_3 \] \[ 1.98 \times \sin(18.8^\circ) = 1.00 \times \sin\theta_3 \] \[ \sin\theta_3 = 1.98 \times 0.322 = 0.6376 \] \[ \theta_3 = \arcsin(0.6376) = 39.6^\circ \] **Explanation:** The angle of incidence at the bottom surface is the same as the angle of refraction at the top surface. When the light exits, the angle returns to the original angle in air. --- ### **Final Answer Summary** **(a) Angle of refraction at the top surface and angle of incidence at the bottom surface:** \( 18.8^\circ \) **(b) Refracted angle at the bottom surface (as it exits):** \( 39.6^\circ \) --- **Next Steps** Would you like the same step-by-step structure for parts (c) through (f)?

Certainly! Continuing with the same detailed, step-by-step approach: --- ## **Part (c): Find the lateral distance \( d \) by which the light beam is shifted** ### **Step 1: Understand the geometry and set up the calculation** **Explanation:** The beam enters the glass at an angle \( \theta_2 = 18.8^\circ \), travels through the thickness of the glass \( t = 0.0200\, \text{m} \). Due to refraction, the beam shifts laterally from its original path. The lateral shift \( d \) can be found using the relation: \[ d = t \times \tan \theta_2 \] This gives the perpendicular displacement of the beam as it passes through the glass. --- ### **Step 2: Calculate the lateral shift \( d \)** \[ d = 0.0200\, \text{m} \times \tan(18.8^\circ) \] Calculate \( \tan(18.8^\circ) \): \[ \tan(18.8^\circ) \approx 0.341 \] Now: \[ d = 0.0200 \times 0.341 = 0.00682\, \text{m} \] **Final value:** \[ \boxed{ d \approx 6.82\, \text{mm} } \] --- ## **Part (d): Calculate the speed of light in the glass** ### **Step 1: Use the relation between speed of light in a medium and refractive index** **Explanation:** The speed of light in a medium is: \[ v = \frac{c}{n} \] where: - \( c = 3.00 \times 10^8\, \text{m/s} \) - \( n = 1.98 \) --- ### **Step 2: Perform the calculation** \[ v = \frac{3.00 \times 10^8}{1.98} \approx 1.52 \times 10^8\, \text{m/s} \] **Final answer:** \[ \boxed{ v \approx 1.52 \times 10^8\, \text{m/s} } \] --- ## **Part (e): Does the travel time through the glass depend on the angle of incidence?** ### **Explanation:** - When light enters at a larger angle, its path length inside the glass increases because the beam travels a longer diagonal distance through the thickness. - The travel time \( t_{travel} = \frac{\text{path length}}{v} \). - With a larger incident angle, the effective path length: \[ L = \frac{t}{\cos \theta_2} \] - Since \( \cos \theta_2 \) decreases as \( \theta_2 \) increases, \( L \) increases, making the travel time longer. **Conclusion:** \[ \boxed{ \text{Yes, a larger incident angle increases the travel time through the glass.} } \] --- ## **Final Summary:** - **(a)** The angle of refraction at the top surface: **18.8°** - **(b)** The refraction angle at the bottom surface: **39.6°** - **(c)** Lateral shift \( d \): **approximately 6.82 mm** - **(d)** Speed of light in glass: **approximately \( 1.52 \times 10^8\, \text{m/s} \)** - **(e)** The travel time through the glass increases with larger incident angles because the light travels a longer path inside the medium. Let me know if you'd like any further clarification!

Certainly! Here's the detailed, step-by-step solution for part (f): --- ## **Part (f): Is the travel time through the glass affected by the angle of incidence?** ### **Step 1: Understand how the path length varies with the angle** **Explanation:** The time taken for light to pass through the glass depends on the path length inside the medium and the speed of light within it. The path length \(L\) inside the glass is longer than the thickness \(t\) when the light enters at an angle \( \theta_2 \): \[ L = \frac{t}{\cos \theta_2} \] As the incident angle increases, \( \theta_2 \) (the refraction angle inside the glass) also increases (from Snell's Law). Since \( \cos \theta_2 \) decreases with larger \( \theta_2 \), the path length \(L\) becomes longer. --- ### **Step 2: Determine how the travel time depends on the path length** The travel time \( T \) inside the glass is: \[ T = \frac{L}{v} \] where \( v \) is the speed of light in glass, calculated earlier as approximately \( 1.52 \times 10^8\, \text{m/s} \). Substituting \( L \): \[ T = \frac{t}{v \cos \theta_2} \] Since \( t \) and \( v \) are constants, the travel time increases as \( \cos \theta_2 \) decreases (i.e., as \( \theta_2 \) increases). Therefore, a larger incident angle results in a longer travel time through the glass. --- ### **Final answer:** **Yes, the travel time through the glass is affected by the angle of incidence.** A larger incident angle causes the light to traverse a longer path inside the glass, increasing the travel time. --- **Summary:** - Increasing the incident angle increases the internal path length. - Longer path length leads to longer travel time inside the glass. - Therefore, the travel time is directly influenced by the angle of incidence. Let me know if you need any further explanation!

Certainly! Let's solve part (e): **Calculate the time required for the light to pass through the glass block**. --- ### **Step 1: Determine the effective path length inside the glass** **Explanation:** The light travels through the glass at an angle \( \theta_2 = 18.8^\circ \) (from previous parts). The physical thickness of the glass is \( t = 0.0200\, \text{m} \). The actual path length \( L \) inside the glass along the beam's trajectory is longer than the thickness because of the angle: \[ L = \frac{t}{\cos \theta_2} \] Calculate \( \cos 18.8^\circ \): \[ \cos 18.8^\circ \approx 0.949 \] Now, compute \( L \): \[ L = \frac{0.0200\, \text{m}}{0.949} \approx 0.02107\, \text{m} \] **Explanation:** The beam travels approximately 21.07 mm inside the glass along its path, longer than the initial thickness due to the angle. --- ### **Step 2: Calculate the travel time \( T \)** **Explanation:** The time taken to traverse this path is: \[ T = \frac{L}{v} \] where: - \( v \) is the speed of light in glass, calculated earlier as approximately \( 1.52 \times 10^8\, \text{m/s} \). Now, substitute: \[ T = \frac{0.02107\, \text{m}}{1.52 \times 10^8\, \text{m/s}} \approx 1.39 \times 10^{-10}\, \text{s} \] or approximately **139 picoseconds**. --- ### **Final answer:** \[ \boxed{ \text{Time to pass through the glass} \approx 1.39 \times 10^{-10}\, \text{s} } \] --- **Summary:** The light takes roughly **139 picoseconds** to pass through the 2 cm thick glass block at the given incident angle. The travel time depends on the effective path length inside the medium, which increases with the incident angle. Let me know if you'd like any further details!

Certainly! Here is a comprehensive summary of all parts (a) through (f): --- ### **Final Answers Summary** **(a) Angle of refraction at the top surface and the angle of incidence at the bottom surface:** - **Inside the glass:** **18.8°** - **As the beam exits into air:** **39.6°** (the original incident angle) **(b) Refracted angle at the bottom surface:** - **39.6°** (same as the initial incident angle in air) **(c) Lateral shift \( d \):** - **Approximately 6.82 mm** - Calculated using \( d = t \times \tan \theta_2 \) **(d) Speed of light in the glass:** - **Approximately \( 1.52 \times 10^8\, \text{m/s} \)** - Derived from \( v = c/n \) **(e) Effect of incident angle on travel time:** - **Yes**, a larger incident angle increases the travel time through the glass because the beam travels a longer path inside due to the increased angle, which increases the effective path length. **(f) Time for light to pass through the glass:** - **Approximately \( 1.39 \times 10^{-10} \) seconds (139 picoseconds)** - Calculated from the effective path length \( L = t / \cos \theta_2 \) and \( T = L / v \) --- Let me know if you'd like further elaboration on any part!

Certainly! Here's a comprehensive summary of all parts (a) through (f): --- ### **Final Answers Summary** **(a) Angle of refraction at the top surface and the angle of incidence at the bottom surface:** - **Refraction angle inside the glass:** **18.8°** - **Angle at which the beam exits into air:** **39.6°** (same as initial incident angle) **(b) Refracted angle at the bottom surface:** - **39.6°** (the same as the initial incident angle in air) **(c) Lateral shift \( d \):** - **Approximately 6.82 mm** - Calculated using \( d = t \times \tan \theta_2 \) **(d) Speed of light in the glass:** - **Approximately \( 1.52 \times 10^8\, \text{m/s} \)** - Derived from \( v = c/n \) **(e) Effect of incident angle on travel time:** - **Yes**, a larger incident angle increases the travel time through the glass because the beam travels a longer path inside due to the increased angle, which increases the effective path length. **(f) Time for light to pass through the glass:** - **Approximately \( 1.39 \times 10^{-10} \) seconds (139 picoseconds)** - Calculated from the effective path length \( L = t / \cos \theta_2 \) and \( T = L / v \) --- Let me know if you'd like further clarification!