Hello I'm having trouble with this physics question: You are examining the loss in intensity as light passes through a fibre optic cable. The input intensity is measured as 55.1 W/m2 and the output intensity as 28.7 W/m2. Calculate the attenuation of the signal in decibels (dB)
So far I calculated
𝛽 𝑑𝐵 = 10log10 * 𝐼1/𝐼2
𝛽 𝑑𝐵 = 10log10 * 55.1/28.7 and got 19.1 is it correct?
db = 10*Log(Io/Iin.) = 10*Log(28.7/55.1)
To calculate the attenuation of the signal in decibels (dB), you need to use the formula:
β dB = 10 log10(I1/I2),
where β dB represents the attenuation in dB, I1 is the input intensity, and I2 is the output intensity.
In your case, the input intensity (I1) is measured as 55.1 W/m², and the output intensity (I2) is measured as 28.7 W/m². Plugging these values into the formula, we get:
β dB = 10 log10(55.1/28.7).
Now, let's calculate it step by step:
β dB = 10 [log10(55.1) - log10(28.7)]
β dB = 10 [1.7412 - 1.4579]
β dB = 10 [0.2833]
β dB = 2.833
So the attenuation of the signal in decibels (dB) is approximately 2.833 dB.
Therefore, your previous calculation of 19.1 dB is incorrect. The correct answer is approximately 2.833 dB.