Step 1: Identify the angles of incidence and refraction from the diagram. The normal line is N-N'. The angle of incidence ($\theta_P$) is the angle between the incident ray (in medium P) and the normal. From the diagram, $\theta_P = 45^\circ$. The angle of refraction ($\theta_Q$) is the angle between the refracted ray (in medium Q) and the normal. From the diagram, $\theta_Q = 30^\circ$. Step 2: Identify the given velocity. The velocity of the wave in medium Q is $v_Q = 2.0 \times 10^{-1} \text{ m s}^{-1}$. Step 3: Apply Snell's Law. Snell's Law states the relationship between the angles and velocities in two different media: $$\frac{\sin \theta_P}{\sin \theta_Q} = \frac{v_P}{v_Q}$$ We need to find $v_P$, the velocity of the wave in medium P. Rearranging the formula: $$v_P = v_Q \frac{\sin \theta_P}{\sin \theta_Q}$$ Step 4: Substitute the known values into Snell's Law. $$v_P = (2.0 \times 10^{-1} \text{ m s}^{-1}) \times \frac{\sin 45^\circ}{\sin 30^\circ}$$ Step 5: Calculate the sine values. $$\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071$$ $$\sin 30^\circ = 0.5$$ Step 6: Perform the calculation. $$v_P = (2.0 \times 10^{-1}) \times \frac{0.7071}{0.5}$$ $$v_P = (0.2) \times 1.4142$$ $$v_P = 0.28284 \text{ m s}^{-1}$$ Step 7: Express the answer in scientific notation and compare with the given options. $$v_P \approx 2.8 \times 10^{-1} \text{ m s}^{-1}$$ This matches option D. The final answer is $\boxed{\text{D: } 2.8 \times 10^{-1} \text{ m s}^{-1}}$.