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Consider an optical communication system operating at λ = 800 nm. Suppose only 1% of the optical source frequency is the available channel bandwidth for optical communication. No. of channels that can be accommodated for transmitting video T.V signals requiring an approximate band width of 4.5 MHz is _______. 

(A) 3.8 × 10+14 

(B) 3.8 × 1012 

(C) 8.4 × 105 

(D) 1.71 × 107

1 Answer

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Best answer

The correct option is (C) 8.4 × 105.

Explanation:

Optical source frequency = f (c/λ) = [(3 × 108) / (800 × 10–9)] = 3.8 × 1014Hz

BW of channel = 1% of f

∴ BW of channel = 3.8 × 1012Hz

No. of channels = [(Total BW of channel) / (BW needed per channel)]

= [(3.8 × 1014) / (4.5 × 106)]

 ∴ No. of channels = 8.4 × 105

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