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This is problem number 40 in chapter 23.
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For this problem, we are given a bat emitting a high -pitch frequency to detect this prey, and we are given the deflection angle to the first dark band.
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The frequency used by the bat is given as 8 .0 times 10 to the 14 hertz.
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And that angle theta is 90 degrees.
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So this angle represents that deflection to the first dark band.
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So the equation that we need to use here is sine of theta is equal to 1 .22 lambda over d.
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We are looking for the diameter of the smallest object that can produce such deflection for this given frequency.
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So the bat is using this frequency to resolve this diameter object.
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So here what we have to do is solve this equation for the diameter of the object.
01:36
We multiply both sides by d.
01:38
We get d sine of theta is equal to 1 .22 lambda and now we divide both sides by sign of that deflection angle.
02:03
So in this problem we aren't given the wavelength, but we are going to be a wavelength.
02:06
But we are going to given the frequency used.
02:09
So this frequency is propagating through air as sound and the speed of sound is 340 meters per second...