Question

Microwaves travel at the speed of light, 3x108m/s. When frequency of the microwaves is 9x109 Hz, quantitatively translate this information into their wavelength and period. Wavelength= 0.03m; period 1.1x10-10s , , Wavelength= 0.03m; period 1.1x10 -10 s, Wavelength= 0.03m; period 1.1x10-10s Wavelength= 0.03m; period 1.1x10 -10 s, Wavelength = 0.03cm; period 1.1x109s Wavelength = 0.03cm; period 1.1x10 9 s, Wavelength=0.03m; period 1.1x1010s

Answer 1

Explanation:

Given that,

The frequency of microwaves, $f=9\times 10^9\ Hz$

The time period of the wave is :

$t=\dfrac{1}{f}\\\\t=\dfrac{1}{9\times 10^9}\\\\t=1.1\times 10^{-10}\ s$

Let $\lambda$ is the wavelength of the wave. We know that,

$v=f\lambda\\\\\lambda=\dfrac{v}{f}$

Put all the values,

$\lambda=\dfrac{3\times 10^8}{9\times 10^{9}}\\\\\lambda=0.03\ m$

Hence, this is the required solution.