Elements of natural philosophy (Volume 2-3) . Illustration; MA sin 9A B sin 9 r> but A M and A B, being described in the same time, Explanation;the first by the incident, the second by the transmittedpulse, are respectively proportional to the velocities inthe two media. Denoting the velocity of the incidentpulse by J7, and that of the transmitted pulse by V\we have whence V MA V AB sin 9 sin 9 Eatio ofvelocities ofincident andrefracted sound; sin 9 sin 9, V (29). That is to say, in the refraction of sound the sine of the of incidence is equal to the sine of the angle of re-fract
Elements of natural philosophy (Volume 2-3) . Illustration; MA sin 9A B sin 9 r> but A M and A B, being described in the same time, Explanation;the first by the incident, the second by the transmittedpulse, are respectively proportional to the velocities inthe two media. Denoting the velocity of the incidentpulse by J7, and that of the transmitted pulse by V\we have whence V MA V AB sin 9 sin 9 Eatio ofvelocities ofincident andrefracted sound; sin 9 sin 9, V (29). That is to say, in the refraction of sound the sine of the of incidence is equal to the sine of the angle of re-fraction multiplied into the ratio obtained by dividingthe velocity before incidence by that after refraction. 32 NATURAL PHILOSOPHY. Application toair and water; Illustration; Thus, if sound proceed through the atmosphere at 32°Fahr., and he incident upon the surface A B, of water at /- the same temperature, then will V = 1089,42, V=4707,4, and F 1089^2 _2V 4707,4 which in Eq. (29) gives sin 9 = 0,23142 . sin 9 or. A^ Example; sm 90,23142 = sm 9 (30). Now, suppose the angle of incidence H IN, to be given,say 30°. With the point of incidence 7, as a centre andradius unity, taken from any scale of equal parts, de-scribe the circumference of a circle; from a table of natu-ral sines take the sine of 30°, and by means of the samescale lay it off from I to H; through H draw II O pa-rallel to the normal NI, and through the point (?, inwhich this parallel meets the circumference and the pointof incidence /, draw li I This gives the incident of Divide the sine of 30° by 0,23142, this will give the sine incident and of / J off fe ya|ue fr()m J t() fit an(J draw J£> Q refracted rays. .. parallel to ly I; join the point in which this parallel cutsthe circumference with the point of incidence 7, and wehave the direction of the refracted ray I Hr. When soundcannot pass intoa secondmedium; § 74. The sine of an angle can never exceed , therefore, the angle of in
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