Question

In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 10⁸ m/s, h = 6.6 × 10⁻³⁴ J-s)
(A) 3.75 × 10⁶
(B) 3.86 × 10⁶
(C) 6.25 × 10⁵
(D) 4.87 × 10⁵

Answer 1

The number of channels accommodated for transmitting TV signals of band width 6 MHz are

(C) 6.25 × 10⁵

Given the velocity of light (c) is 3 × 10⁸ m/s.

Again,the wavelength (λ) = 800 nm = 8 × 10⁻⁷ m        [10⁻⁹ m]

We know that , frequency (f) is =

c/λ = (3 × 10⁸ / 8 × 10⁻⁷) Hz

=  3.75 × 10¹⁴ Hz

Given that, 1% of frequency is available for bandwidth.

So, 1% of f is = 1/100 * f = 3.75 × 10¹² Hz

= 3.75 × 10⁶ MHz                                    [ 1MHz = 10⁶ Hz]

Number of bands of width = 6MHz

= 3.75 × 10⁶/6 MHz  

= 6.25 × 10⁵ MHz

Option (C) is correct.