The intensity of light pulse travelling in an optical fiber decreases 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) In\({\left(\frac{{{1}}}{{\alpha}}\right)}\)
(B) In\({\left(\alpha\right)}\)
(C) \(\frac{{{\left(\text{In}{5}\right)}}}{{\alpha}}\)
(D) In \({\left(\frac{{{5}}}{{\alpha}}\right)}\)
(A) In\({\left(\frac{{{1}}}{{\alpha}}\right)}\)
(B) In\({\left(\alpha\right)}\)
(C) \(\frac{{{\left(\text{In}{5}\right)}}}{{\alpha}}\)
(D) In \({\left(\frac{{{5}}}{{\alpha}}\right)}\)
Answer
Answer: C
\(\frac{{{I}}}{{{I}_{{{0}}}}}={e}^{{-\alpha{x}}}=\frac{{{20}}}{{{100}}}={e}^{{\alpha{x}}}={5}\)
\({a}{x}={{\log}_{{{e}}}{5}},{x}=\frac{{{\left(\text{In}{5}\right)}}}{{\alpha}}\)
OR
\({I}={I}_{{{0}}}{e}^{{-\alpha{x}}}\)
\({0.2}{I}_{{{0}}}={I}_{{{0}}}{e}^{{-\alpha{x}}}\)
\(\frac{{{2}}}{{{10}}}={e}^{{-\alpha{x}}}\Rightarrow\frac{{{10}}}{{{2}}}={e}^{{\alpha{x}}}\)
\({l}{n}{\left({5}\right)}={\left.{d}{x}\right.}\)
\({x}=\frac{{{l}{n}{\left({5}\right)}{J}}}{{\alpha}}\)
\(\frac{{{I}}}{{{I}_{{{0}}}}}={e}^{{-\alpha{x}}}=\frac{{{20}}}{{{100}}}={e}^{{\alpha{x}}}={5}\)
\({a}{x}={{\log}_{{{e}}}{5}},{x}=\frac{{{\left(\text{In}{5}\right)}}}{{\alpha}}\)
OR
\({I}={I}_{{{0}}}{e}^{{-\alpha{x}}}\)
\({0.2}{I}_{{{0}}}={I}_{{{0}}}{e}^{{-\alpha{x}}}\)
\(\frac{{{2}}}{{{10}}}={e}^{{-\alpha{x}}}\Rightarrow\frac{{{10}}}{{{2}}}={e}^{{\alpha{x}}}\)
\({l}{n}{\left({5}\right)}={\left.{d}{x}\right.}\)
\({x}=\frac{{{l}{n}{\left({5}\right)}{J}}}{{\alpha}}\)
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