Edit to “Solution”
en/3.7.14.md
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| ### Statement | |||
| $3.7.14.$ A plane wave is incident on a plane interface with sound velocity $c_1$ at angle $\alpha$ to the normal. Find the direction of propagation of the reflected and refracted wave if the velocity of the wave in the second medium is equal to $c_2$. | |||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | |||
| − |  | ||
| The blue lines in the figure are the same wavefront at different equally-spaced time points. The incident angle is $\angle DAB=\alpha$. When the wavefront reaches 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 | |||
| \[\angle FCB=\arcsin\left(\frac{c_2}{c_1}\cdot\sin\alpha\right).\] | |||
| #### Answer | |||
| Reflected angle = $\alpha$. | |||
| Refracted angle = $\arcsin\left(\frac{c_2}{c_1}\cdot\sin\alpha\right)$. | |||
| unchanged lines 13 | |||
| ### Statement | ### Statement | ||
| $3.7.14.$ A plane wave is incident on a plane interface with sound velocity $c_1$ at angle $\alpha$ to the normal. Find the direction of propagation of the reflected and refracted wave if the velocity of the wave in the second medium is equal to $c_2$. | $3.7.14.$ A plane wave is incident on a plane interface with sound velocity $c_1$ at angle $\alpha$ to the normal. Find the direction of propagation of the reflected and refracted wave if the velocity of the wave in the second medium is equal to $c_2$. | ||
| @@ -4,7 +4,7 @@Statement | |||
| ### Solution | ### Solution | ||
|  | ||
| The blue lines in the figure are the same wavefront at different equally-spaced time points. The incident angle is $\angle DAB=\alpha$. When the wavefront reaches 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 blue lines in the figure are the same wavefront at different equally-spaced time points. The incident angle is $\angle DAB=\alpha$. When the wavefront reaches 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, | The ratio of the radii of the red and green spherical waves originating from $B$ is $c_1/c_2$. Consequently, | ||
| \[\frac{\sin\angle ECB}{\sin\angle FCB}=\frac{BE/BC}{BF/BC}=\frac{c_1}{c_2}.\] | \[\frac{\sin\angle ECB}{\sin\angle FCB}=\frac{BE/BC}{BF/BC}=\frac{c_1}{c_2}.\] | ||
| Therefore, the refracted angle is | Therefore, the refracted angle is | ||
| \[\angle FCB=\arcsin\left(\frac{c_2}{c_1}\cdot\sin\alpha\right).\] | \[\angle FCB=\arcsin\left(\frac{c_2}{c_1}\cdot\sin\alpha\right).\] | ||
| #### Answer | #### Answer | ||
| Reflected angle = $\alpha$. | Reflected angle = $\alpha$. | ||
| Refracted angle = $\arcsin\left(\frac{c_2}{c_1}\cdot\sin\alpha\right)$. | Refracted angle = $\arcsin\left(\frac{c_2}{c_1}\cdot\sin\alpha\right)$. | ||
| unchanged lines 13 | |||