Question

dvanced photodiode detectors have a second light-emitting diode, operating at a wavelength of 2.0 × 10–7 m, to detect even smaller smoke particles from smoldering flames. What is the frequency difference between the two light beams?

Answer 1

`1.5\cdot 10^(15)Hz`

Step-by-step explanation:

The relationship between wavelength and frequency is given by:

`c=f \lambda`

where

`c=3\cdot 10^8 m/s` is the speed of light

f is the frequency

`\lambda` is the wavelength

In this problem, we know the wavelength:

`\lambda=2.0\cdot 10^(-7) m`

So we can re-arrange the formula above to find the frequency:

`f=(c)/(\lambda)=(3.0\cdot 10^8 m/s)/(2.0\cdot 10^(-7) m)=1.5\cdot 10^(15)Hz`