Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission?

Question

In a microscope of tube length $10\,\text{cm}$ two convex lenses are arranged with focal lengths $2\,\text{cm}$ and $5\,\text{cm}$. Total magnification obtained with this system for normal adjustment is $(5)^k$. The value of $k$ is ___.

Answer 1

The problem involves determining how many audio signal channels can be accommodated in an optical communication system given certain specifications. Let's break down the calculations step-by-step.

First, identify the optical source frequency corresponding to the given wavelength of 1000 nm. The relationship between frequency (f) and wavelength (\lambda) is given by the equation: c = f \lambda, where c = 3 \times 10^8 \, \text{m/s}, the speed of light.
Rearranging the formula for frequency, we have: f = \frac{c}{\lambda}. Plugging in the values: f = \frac{3 \times 10^8}{1000 \times 10^{-9}} which gives f = 3 \times 10^{14} \, \text{Hz}.
The available bandwidth is 2% of this optical source frequency. Therefore, the channel bandwidth is: \text{Bandwidth} = 0.02 \times 3 \times 10^{14} = 6 \times 10^{12} \, \text{Hz}.
Each audio signal requires a bandwidth of 8 kHz, or 8 \times 10^3 \, \text{Hz}.
To find the number of channels N that can fit into the available bandwidth, divide the total bandwidth by the bandwidth per channel: N = \frac{\text{Total Bandwidth}}{\text{Bandwidth per Channel}}. Plug in the values: N = \frac{6 \times 10^{12}}{8 \times 10^3} = 7.5 \times 10^8.

To express this result in the options format, we rewrite 7.5 \times 10^{8} as 75 \times 10^{7}. Thus, the correct answer is 75 × 10⁷.

This calculation confirms that the optical bandwidth can accommodate 75 × 10⁷ audio signal channels, matching the given correct answer.

Answer 2

The problem involves determining how many audio signal channels can be accommodated in an optical communication system given certain specifications. Let's break down the calculations step-by-step.

First, identify the optical source frequency corresponding to the given wavelength of 1000 nm. The relationship between frequency (f) and wavelength (\lambda) is given by the equation: c = f \lambda, where c = 3 \times 10^8 \, \text{m/s}, the speed of light.
Rearranging the formula for frequency, we have: f = \frac{c}{\lambda}. Plugging in the values: f = \frac{3 \times 10^8}{1000 \times 10^{-9}} which gives f = 3 \times 10^{14} \, \text{Hz}.
The available bandwidth is 2% of this optical source frequency. Therefore, the channel bandwidth is: \text{Bandwidth} = 0.02 \times 3 \times 10^{14} = 6 \times 10^{12} \, \text{Hz}.
Each audio signal requires a bandwidth of 8 kHz, or 8 \times 10^3 \, \text{Hz}.
To find the number of channels N that can fit into the available bandwidth, divide the total bandwidth by the bandwidth per channel: N = \frac{\text{Total Bandwidth}}{\text{Bandwidth per Channel}}. Plug in the values: N = \frac{6 \times 10^{12}}{8 \times 10^3} = 7.5 \times 10^8.

To express this result in the options format, we rewrite 7.5 \times 10^{8} as 75 \times 10^{7}. Thus, the correct answer is 75 × 10⁷.

This calculation confirms that the optical bandwidth can accommodate 75 × 10⁷ audio signal channels, matching the given correct answer.

Only 2% of the optical source frequency is the available channel bandwidth for an optical communicating system operating at 1000 nm. If an audio signal requires a bandwidth of 8 kHz, how many channels can be accommodated for transmission?

The Correct Option is B

Solution and Explanation

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