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

Answer:

$1.5\cdot 10^{15}Hz$

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:

$f=\frac{c}{\lambda}=\frac{3.0\cdot 10^8 m/s}{2.0\cdot 10^{-7} m}=1.5\cdot 10^{15}Hz$