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en/3.7.14.md
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| + | ### Statement | ||
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| + | $3.7.14.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + |  | ||
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| + | The blue lines in the figure are the same wavefront at different equally-spaced time points. The angle of incident is $\angle DAB=\alpha$. When the wavefront reach the interface of the two media at $A$, $B$, $C$, these points become sources of new spherical waves. (The figure shows the moment when the wavefront has just reached $C$.) Since the red spherical wave originating from $B$ is tangent to the first blue line and the reflected wavefront (as the speed of the wavefront equals the speed of sound), we have $\triangle DAB\cong\triangle ECB$. Thus, the reflected angle $\angle ECB=\angle DAB=\alpha$. | ||
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| + | The ratio of the radii of the red and green spherical waves originating from $B$ is $c_1/c_2$. Consequently, | ||
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| + | \[\frac{\sin\angle ECB}{\sin\angle FCB}=\frac{BE/BC}{BF/BC}=\frac{c_1}{c_2}.\] | ||
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| + | Therefore, the refracted angle is | ||
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| + | \[\angle FCB=\arcsin\left(\frac{c_2}{c_1}\sin\alpha\right).\] | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.7.14.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| The blue lines in the figure are the same wavefront at different equally-spaced time points. The angle of incident is $\angle DAB=\alpha$. When the wavefront reach the interface of the two media at $A$, $B$, $C$, these points become sources of new spherical waves. (The figure shows the moment when the wavefront has just reached $C$.) Since the red spherical wave originating from $B$ is tangent to the first blue line and the reflected wavefront (as the speed of the wavefront equals the speed of sound), we have $\triangle DAB\cong\triangle ECB$. Thus, the reflected angle $\angle ECB=\angle DAB=\alpha$. | |||
| The ratio of the radii of the red and green spherical waves originating from $B$ is $c_1/c_2$. Consequently, | |||
| \[\frac{\sin\angle ECB}{\sin\angle FCB}=\frac{BE/BC}{BF/BC}=\frac{c_1}{c_2}.\] | |||
| Therefore, the refracted angle is | |||
| \[\angle FCB=\arcsin\left(\frac{c_2}{c_1}\sin\alpha\right).\] | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||