The intensity of a light pulse travelling along an optical fibre decreases exponentially with distance according to the relation `l = i_(0) e^(-0.0693x)` where x is in km and `l_(0)` is intensity of incident pulse. The intensity of pulse reduces to `(1)/(4)` after travelling a distance

Options

A. 1km
B. 10 km
C. 20 km
D. 40 km

Answer

To solve the problem, we start with the given relation for the intensity of a light pulse traveling along an optical fiber:

\[ I = I_0 e^{-0.0693x} \]

where:
- \( I \) is the intensity after traveling a distance \( x \) (in km),
- \( I_0 \) is the initial intensity,
- \( x \) is the distance traveled in kilometers.

We need to find the distance \( x \) at which the intensity \( I \) reduces to \( \frac{1}{4} I_0 \).

### Step 1: Set up the equation

We know that the intensity reduces to \( \frac{1}{4} I_0 \). Therefore, we can set up the equation:

\[ \frac{1}{4} I_0 = I_0 e^{-0.0693x} \]

### Step 2: Simplify the equation

We can divide both sides of the equation by \( I_0 \) (assuming \( I_0 \neq 0 \)):

\[ \frac{1}{4} = e^{-0.0693x} \]

### Step 3: Take the natural logarithm of both sides

To solve for \( x \), we take the natural logarithm (ln) of both sides:

\[ \ln\left(\frac{1}{4}\right) = -0.0693x \]

### Step 4: Solve for \( x \)

Now, we can isolate \( x \):

\[ x = \frac{\ln\left(\frac{1}{4}\right)}{-0.0693} \]

### Step 5: Calculate \( \ln\left(\frac{1}{4}\right) \)

We know that:

\[ \ln\left(\frac{1}{4}\right) = \ln(1) - \ln(4) = 0 - \ln(4) = -\ln(4) \]

Using the property of logarithms, we can also express \( \ln(4) \) as:

\[ \ln(4) = \ln(2^2) = 2\ln(2) \]

Thus,

\[ \ln\left(\frac{1}{4}\right) = -2\ln(2) \]

### Step 6: Substitute back into the equation

Substituting this back into our equation for \( x \):

\[ x = \frac{-2\ln(2)}{-0.0693} \]

### Step 7: Calculate \( x \)

Now we can compute the value of \( x \):

\[ x = \frac{2\ln(2)}{0.0693} \]

Using the approximate value \( \ln(2) \approx 0.693 \):

\[ x \approx \frac{2 \times 0.693}{0.0693} \]

Calculating this gives:

\[ x \approx \frac{1.386}{0.0693} \approx 20 \text{ km} \]

### Final Answer

The distance \( x \) at which the intensity reduces to \( \frac{1}{4} I_0 \) is approximately **20 km**.

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