Estimate the fastest bit rate capable of being carried by light of wavelength `1.3 mu m`. How many phone calls could be carried at this bit rate ? Band width of optical fibre = 2 GHz.

Options

A. `2.9xx10^(7)`
B. `1.5xx10^(6)`
C. `2.3xx10^(5)`
D. `1.7xx10^(4)`

Explanation

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Calculate the Frequency of Light
We are given the wavelength of light, \( \lambda = 1.3 \, \mu m = 1.3 \times 10^{-6} \, m \). We can find the frequency \( \nu \) using the formula:

\[ \nu = \frac{c}{\lambda} \]

where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, m/s \).

Substituting the values:

\[ \nu = \frac{3 \times 10^8 \, m/s}{1.3 \times 10^{-6} \, m} \approx 2.31 \times 10^{14} \, Hz \]

### Step 2: Calculate the Maximum Bit Rate
The maximum bit rate \( R \) can be calculated using the formula:

\[ R = 2 \times \nu \]

Substituting the frequency we found in Step 1:

\[ R = 2 \times 2.31 \times 10^{14} \, Hz \approx 4.62 \times 10^{14} \, bps \]

### Step 3: Determine the Bandwidth of the Optical Fiber
The bandwidth of the optical fiber is given as \( 2 \, GHz \). We convert this to hertz:

\[ \text{Bandwidth} = 2 \, GHz = 2 \times 10^9 \, Hz \]

### Step 4: Calculate the Number of Phone Calls
To find the number of phone calls that can be carried at this bit rate, we use the formula:

\[ \text{Number of phone calls} = \frac{R}{\text{Bandwidth}} \]

Substituting the values we have:

\[ \text{Number of phone calls} = \frac{4.62 \times 10^{14} \, bps}{2 \times 10^9 \, Hz} \approx 2.31 \times 10^5 \]

### Final Answer
The fastest bit rate capable of being carried by light of wavelength \( 1.3 \, \mu m \) is approximately \( 4.62 \times 10^{14} \, bps \), and the number of phone calls that could be carried at this bit rate is approximately \( 2.31 \times 10^5 \).

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