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Identify the angles of incidence and refraction from the diagram.

Physics

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Answer

D: 2.8×10−1 m s−1\text{D: } 2.8 \times 10^{-1} \text{ m s}^{-1}

Step-by-step solution

7 steps

  1. Identify the angles of incidence and refraction from the diagram.

    The normal line is N-N'.
    The angle of incidence (θP\theta_P) is the angle between the incident ray (in medium P) and the normal. From the diagram, θP=45∘\theta_P = 45^\circ.
    The angle of refraction (θQ\theta_Q) is the angle between the refracted ray (in medium Q) and the normal. From the diagram, θQ=30∘\theta_Q = 30^\circ.

  2. Identify the given velocity.

    The velocity of the wave in medium Q is vQ=2.0×10−1 m s−1v_Q = 2.0 \times 10^{-1} \text{ m s}^{-1}.

  3. Apply Snell's Law.

    Snell's Law states the relationship between the angles and velocities in two different media:

    sin⁡θPsin⁡θQ=vPvQ\frac{\sin \theta_P}{\sin \theta_Q} = \frac{v_P}{v_Q}

    We need to find vPv_P, the velocity of the wave in medium P. Rearranging the formula:

    vP=vQsin⁡θPsin⁡θQv_P = v_Q \frac{\sin \theta_P}{\sin \theta_Q}

  4. Substitute the known values into Snell's Law.

    vP=(2.0×10−1 m s−1)×sin⁡45∘sin⁡30∘v_P = (2.0 \times 10^{-1} \text{ m s}^{-1}) \times \frac{\sin 45^\circ}{\sin 30^\circ}

  5. Calculate the sine values.

    sin⁡45∘=22≈0.7071\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071 sin⁡30∘=0.5\sin 30^\circ = 0.5

  6. Perform the calculation.

    vP=(2.0×10−1)×0.70710.5v_P = (2.0 \times 10^{-1}) \times \frac{0.7071}{0.5} vP=(0.2)×1.4142v_P = (0.2) \times 1.4142 vP=0.28284 m s−1v_P = 0.28284 \text{ m s}^{-1}

  7. Express the answer in scientific notation and compare with the given options.

    vP≈2.8×10−1 m s−1v_P \approx 2.8 \times 10^{-1} \text{ m s}^{-1}

    This matches option D.

    The final answer is D: 2.8×10−1 m s−1\text{D: } 2.8 \times 10^{-1} \text{ m s}^{-1}.

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