QuestionJuly 4, 2025

Suppose that the propagation velocity of Medium 2 is 1300m/sec and the propagation velocity of Medium 1 is 1200m/sec For an incident angle is 30^circ is the refracted angle toward or away from normal? A away from B toward C no refraction occurs

Suppose that the propagation velocity of Medium 2 is 1300m/sec and the propagation velocity of Medium 1 is 1200m/sec For an incident angle is 30^circ is the refracted angle toward or away from normal? A away from B toward C no refraction occurs
Suppose that the propagation velocity of Medium 2 is 1300m/sec and the propagation velocity of
Medium 1 is 1200m/sec For an incident angle is 30^circ  is the refracted angle toward or away from normal?
A away from
B toward
C no refraction occurs

Solution
4.5(181 votes)

Answer

(B) toward Explanation 1. Apply Snell's Law Use **Snell's Law**: n_1 \sin(\theta_1) = n_2 \sin(\theta_2), where n_1 and n_2 are the refractive indices. Since n = \frac{c}{v}, we have \frac{v_1}{v_2} = \frac{\sin(\theta_2)}{\sin(\theta_1)}. Given v_1 = 1200 \mathrm{~m/s}, v_2 = 1300 \mathrm{~m/s}, and \theta_1 = 30^{\circ}. 2. Calculate \sin(\theta_2) \sin(\theta_2) = \frac{v_1}{v_2} \cdot \sin(30^{\circ}) = \frac{1200}{1300} \cdot 0.5 = 0.4615 3. Determine \theta_2 \theta_2 = \arcsin(0.4615) \approx 27.6^{\circ} 4. Compare angles Since \theta_2 < \theta_1, the refracted angle is toward the normal.

Explanation

1. Apply Snell's Law<br /> Use **Snell's Law**: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$, where $n_1$ and $n_2$ are the refractive indices. Since $n = \frac{c}{v}$, we have $\frac{v_1}{v_2} = \frac{\sin(\theta_2)}{\sin(\theta_1)}$. Given $v_1 = 1200 \mathrm{~m/s}$, $v_2 = 1300 \mathrm{~m/s}$, and $\theta_1 = 30^{\circ}$.<br />2. Calculate $\sin(\theta_2)$<br /> $\sin(\theta_2) = \frac{v_1}{v_2} \cdot \sin(30^{\circ}) = \frac{1200}{1300} \cdot 0.5 = 0.4615$<br />3. Determine $\theta_2$<br /> $\theta_2 = \arcsin(0.4615) \approx 27.6^{\circ}$<br />4. Compare angles<br /> Since $\theta_2 < \theta_1$, the refracted angle is toward the normal.
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