If the distance to a reflector doubles, what happens to the time between pulse transmission and echo detection?

When pulse transmission occurs in ultrasound, the time it takes for the sound waves to travel to a reflector and back is closely related to the distance that needs to be covered. When the distance to the reflector is doubled, the sound wave must travel twice the original distance to reach the reflector and then return back to the transducer.

To understand this concept more clearly, consider that the speed of sound in tissue is relatively constant at approximately 1540 meters per second. The time taken for the sound wave to travel a given distance can be calculated using the formula: time = distance/speed.

If the original distance to the reflector is (d), the round trip (to the reflector and back) takes time (t_1), which can be represented as:

[ t_1 = \frac{2d}{1540} ]

If the distance is doubled to (2d), the new round trip time (t_2) becomes:

[ t_2 = \frac{2(2d)}{1540} = \frac{4d}{1540} ]

From this, we can see that (t_2) is double (t_1) since ( \frac{4