The intensity of a light pulse travelling alo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The attenuation in decibels $(dB)$ is given by the formula: $10 \log_{10} (I/I_0) = -\alpha_{dB} x$,where $\alpha_{dB}$ is the attenuation constan

Vedclass · Accepted Answer

$0.0602$

Vedclass · Answer

(B) The attenuation in decibels $(dB)$ is given by the formula: $10 \log_{10} (I/I_0) = -\alpha_{dB} x$,where $\alpha_{dB}$ is the attenuation constant in $dB/km$.Given that the intensity falls by $50$ percent,the final intensity $I = 0.5 I_0$,which means $I/I_0 = 1/2$.The distance $x = 50 \ km$.Substituting these values into the formula:$10 \log_{10} (1/2) = -\alpha_{dB} 	imes 50$$10 (\log_{10} 1 - \log_{10} 2) = -50 \alpha_{dB}$Since $\log_{10} 1 = 0$ and $\log_{10} 2 \approx 0.3010$:$10 (0 - 0.3010) = -50 \alpha_{dB}$$-3.010 = -50 \alpha_{dB}$$\alpha_{dB} = 3.010 / 50 = 0.0602 \ dB/km$.

Similar Questions

The relationship between the maximum electron density $(N_{max})$ of the ionosphere and the critical frequency $(f_c)$ is given by:

Audio signals cannot be transmitted directly over long distances because:

$A$ message signal of frequency $8 \text{ kHz}$ and peak voltage $12 \text{ V}$ is used to modulate a carrier of frequency $1.2 \text{ MHz}$ and peak voltage $20 \text{ V}$. The modulation index is

Television signals broadcast from the moon can be received on the earth,while the $TV$ broadcast from Delhi cannot be received at places about $100 \, km$ distant from Delhi. This is because:

In the case of amplitude modulation,to avoid distortion,the modulation index $(\mu)$ should be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module