Правка раздела «Solution»
en/3.7.16.md
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| ### Statement | |||
| $3.7.16.$ When a plane sound wave falls rather flat on the interface between two media from a medium in which the speed of sound is higher, no refracted wave is formed in the second medium. This phenomenon is called total internal reflection. Find the angle of total internal reflection if the speed of sound in these media is equal to $c_1$ and $c_2$ ($c_1<c_2$). | |||
| @@ -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 $C$.) If the incident angle is critical, then the green spherical waves in the second medium are internally tangent and no envelope is formed. Since the red spherical wave originating from $B$ is tangent to the first blue line (as the speed of the wavefront equals the speed of sound) and the ratio of the radii of the red and green spherical waves is $c_1/c_2$, we have $\sin\alpha=BD/BA=c_1/c_2$. Thus, the critical angle is $\arcsin(c_1/c_2)$. | |||
| #### Answer | |||
| $\arcsin(c_1/c_2)$ | |||
| ещё строк без изменений 3 | |||
| ### Statement | ### Statement | ||
| $3.7.16.$ When a plane sound wave falls rather flat on the interface between two media from a medium in which the speed of sound is higher, no refracted wave is formed in the second medium. This phenomenon is called total internal reflection. Find the angle of total internal reflection if the speed of sound in these media is equal to $c_1$ and $c_2$ ($c_1<c_2$). | $3.7.16.$ When a plane sound wave falls rather flat on the interface between two media from a medium in which the speed of sound is higher, no refracted wave is formed in the second medium. This phenomenon is called total internal reflection. Find the angle of total internal reflection if the speed of sound in these media is equal to $c_1$ and $c_2$ ($c_1<c_2$). | ||
| @@ -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 $C$.) If the incident angle is critical, then the green spherical waves in the second medium are internally tangent and no envelope is formed. Since the red spherical wave originating from $B$ is tangent to the first blue line (as the speed of the wavefront equals the speed of sound) and the ratio of the radii of the red and green spherical waves is $c_1/c_2$, we have $\sin\alpha=BD/BA=c_1/c_2$. Thus, the critical angle is $\arcsin(c_1/c_2)$. | 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 $C$.) If the incident angle is critical, then the green spherical waves in the second medium are internally tangent and no envelope is formed. Since the red spherical wave originating from $B$ is tangent to the first blue line (as the speed of the wavefront equals the speed of sound) and the ratio of the radii of the red and green spherical waves is $c_1/c_2$, we have $\sin\alpha=BD/BA=c_1/c_2$. Thus, the critical angle is $\arcsin(c_1/c_2)$. | ||
| #### Answer | #### Answer | ||
| $\arcsin(c_1/c_2)$ | $\arcsin(c_1/c_2)$ | ||
| ещё строк без изменений 3 | |||