Question

The intensity of light pulse travelling in an optical fiber decrease according to the relation $I=I_0{e}^{-\alpha x}$. The intensity of light is reduced to $20\%$of its initial value after a distance x equal to

• A. $\log \left(\frac{1}{\alpha }\right)$
• B. $\log \left(\alpha \right)$
• C. $\frac{\left(\log \left(5\right)\right)}{\alpha }$
• D. $\log \left(\frac{5}{\alpha }\right)$

Answer 1

The correct option is C

$\frac{\left(\log \left(5\right)\right)}{\alpha }$

Step - 1: Given

Intensity of light pulse travelling in an optical fiber decrease according to the relation $I=I_0{e}^{-\alpha x}$

Consider initial intensity = $I$

Reduced intensity to $20\%$ as $I_0=\frac{I}{100}\times 20=\frac{I}{5}$

Step - 2: Relation

$I=I_0{e}^{-\alpha x}$

Step - 3: To find the value of x

$$\begin{gathered} I=I_0{e}^{-\alpha x} \\ \Rightarrow \frac{I}{I_0}=e^{-\alpha x} \end{gathered}$$

Applying logarithm on both sides,

$$\begin{gathered} \log \left(\frac{I}{I_0}\right)=\log \left(e^{-\alpha x}\right) \\ \Rightarrow \log \left(\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}\right)=-\alpha x \\ \Rightarrow x=\frac{\log \left(5\right)}{\alpha } \end{gathered}$$

Hence option C is correct.