Question

An ultrasound beam traversing through 5 cm of a material is attenuated at a rate of 2 dB/cm. The intensity of the beam at the exit of the material is \(I_{exit}\) and the intensity at the entry into the material is \(I_{entry}\).
The ratio \(\dfrac{I_{exit}}{I_{entry}} = \)

• A. 0.5
• B. 0.9
• C. 0.1
• D. 0.7

Hint:

Total attenuation = 2 dB/cm x 5 cm = 10 dB; ratio = 10^(-dB/10) = 0.1.

Answer 1

The Correct Option is C

Step 1: Find the total attenuation over the path.
The beam loses \(2\) dB for every centimeter it travels. Over a path length of \(5\) cm, the total attenuation is
\[ \text{Total loss} = 2 \ \text{dB/cm} \times 5 \ \text{cm} = 10 \ \text{dB} \]

Step 2: Recall what a negative dB value means for an intensity ratio.
Attenuation is a loss, so the exit intensity is smaller than the entry intensity. In decibels, this loss is written as
\[ 10\log_{10}\left(\frac{I_{exit}}{I_{entry}}\right) = -10 \ \text{dB} \]
The negative sign shows the ratio is a fraction less than \(1\).

Step 3: Solve for the ratio.
\[ \log_{10}\left(\frac{I_{exit}}{I_{entry}}\right) = \frac{-10}{10} = -1 \]
\[ \frac{I_{exit}}{I_{entry}} = 10^{-1} = 0.1 \]

Step 4: Why the other options are wrong.
Option (A) \(0.5\) would correspond to about \(3\) dB of loss, the loss from only about \(1.5\) cm of travel, not the full \(5\) cm.
Option (B) \(0.9\) corresponds to under \(1\) dB of loss, far too little for \(10\) dB of attenuation.
Option (D) \(0.7\) corresponds to roughly \(3\) dB, again much less than \(10\) dB.

Final Answer:
The ratio \(I_{exit}/I_{entry}\) is \(0.1\).
\[ \boxed{0.1} \]