Question

For any type of wave that reaches a boundary beyond which its speed is increased, there is a maximum incident angle if there is to be a transmitted refracted wave. This maximum incident angle corresponds to an angle of refraction equal to 90°. If all the wave is reflected at the boundary and none is refracted, because refraction would correspond to (where is the angle of refraction), which is impossible. This phenomenon is referred to as total internal reflection. (a) Find a formula for using the law of refraction, Eq. 11–20. (b) How far from the bank should a trout fisherman stand (Fig. 11–61) so trout won’t be frightened by his voice (1.8 m above the ground)? The speed of sound is about 343 m/s in air and 1440 m/s in water.

Short answer

(a) The expression to find the maximum incident angle is \({\theta _{iM}} = \;{\sin ^{ - 1}}\left( {\frac{{{v_a}}}{{{v_w}}}} \right)\).

(b) The fisherman should stand at a distance of 0.44 m from the bank.

Step-by-step solution

Given data

The speed of the sound in the air is \({v_a} = \;343\;{\rm{m/s}}\).

The speed of the sound in the water is \({v_w} = \;1440\;{\rm{m/s}}\).

The distance of the mouth of the fisherman from the ground is \(d\; = \;1.8\;{\rm{m}}\).

Concepts

The law of refraction is given by,

\(\frac{{\sin \;{\theta _i}}}{{\sin \;{\theta _r}}} = \;\frac{{{v_i}}}{{{v_r}}}\)

Here \({\theta _i}\) is the angle of incidence, \({\theta _r}\) is the angle of refraction, \({v_i}\) is the velocity of the wave in the incident medium, and \({\theta _r}\) is the velocity of the wave in the refracted medium.

Calculations

Part (a)

In this case, the incident medium is air, and the refracting medium is water. Therefore, we can rewrite the law of refraction as,

\(\frac{{\sin \;{\theta _i}}}{{\sin \;{\theta _r}}} = \;\frac{{{v_a}}}{{{v_w}}}\)

Now, when the incident angle is maximum, the refracted angle is \(90^\circ \). Substituting this condition to the above equation, we get

\(\begin{aligned}{c}\frac{{\sin \;{\theta _{iM}}}}{{\sin \;90^\circ }} = \;\frac{{{v_a}}}{{{v_w}}}\\\frac{{\sin \;{\theta _{iM}}}}{1} = \;\frac{{{v_a}}}{{{v_w}}}\\\sin \;{\theta _{iM}} = \;\frac{{{v_a}}}{{{v_w}}}\end{aligned}\)

Solving further, we get

\({\theta _{iM}} = \;{\sin ^{ - 1}}\left( {\frac{{{v_a}}}{{{v_w}}}} \right)\)

Therefore, the above expression can be used to find the maximum incident angle.

Part (b)

Substituting the values in the expression for maximum incident angle, we get

\(\begin{aligned}{c}{\theta _{iM}} &= \;{\sin ^{ - 1}}\left( {\frac{{343\;{\rm{m/s}}}}{{1440\;{\rm{m/s}}}}} \right)\\ &= \;13.78^\circ \end{aligned}\)

Let x be the distance that of the fisherman from the bank. Now, from the figure in the question, we get

\(\begin{aligned}{c}\tan {\theta _{iM}} &= \;\frac{x}{d}\\x\; &= \;d\tan {\theta _{iM}}\\ &= \;1.8\;{\rm{m}} \times \;\tan 13.78^\circ \\ &= \;0.44\;{\rm{m}}\end{aligned}\)

Therefore, the fisherman should stand at a distance of 0.44 m from the bank to avoid frightening the trout.