Question

A depth finder sends a 11.4 kHz pulse into an estuary. In this estuary a layer of freshwater (vfw = 1,429.8 m/s), at the surface, rests on top of a layer of saltwater (vsw = 1,556.5 m/s). The pulse travels downward at 2.3° to the vertical, until it encounters the saltwater layer.

(a) Calculate the direction of the pulse in the saltwater.     °

(b) Calculate the critical angle going from fresh to saltwater.     °

(c) A faint echo is heard 0.101 s, after the pulse hits the fresh / saltwater boundary. Calculate the depth of the freshwater.     m

(d) A strong echo from the estuary floor is heard 0.1935 s, after the pulse was sent. Calculate the depth to the bottom of the estuary.     m

Answer 1

(a) the direction of pulse is along from the vertical. Therefore by snell's law

(b)critical angle

(c) If say depth of water is d, echo is heard after the vertical distance covered by pulse(back and forth) is 2d. This distance is covered only by the vertical component of velocity.

(d) time travelled only in the saltwater region = 0.1935 - 0.101 = 0.0925 s
Now as above, depth of seawater region is calculated as

Therefore depth till estuary = 72.1467 + 71.9196 = 144.066 m