Question

An ultrasound equipment installed at a height of 80 m above sea water surface sends out ultrasound that returns from the sea bed and is detected after 5.4 m/s. if the speed of ultrasound through sea water is 1500 m/s and that through air is 320 m/s, then the depth of sea at that place is ?

Answer 1

First of all this question can be solved if we consider normal incidence of the ultrasonic waves instead of oblique incidence because incase of oblique incidence we would require the angle made by ultrasonic waves with the normal at the interface

Time taken from transmission to reception is given as 5.4 seconds (total time)

Therefore,

5.4= time required to travel in water +time required to travel in air

We know time = Distance /Speed

And the distance above sea level is given to be 80 m

Therefore by data,

5.4= (80*2)/320+(d*2)/1500

Where d is the depth of water

2.7=80/320+d/1500

d=(2.7-80/320)*1500

d=2.45*1500

d=3675

Hence the depth of water is 3675 meters

Hope this helps :)