(a) Calculate the ratio of the highest to lowest frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to 760 nm . (b) Compare this with the ratio of highest to lowest frequencies the ear can hear.

Electromagnetic waves are waves that are produced by changes in the electric and magnetic fields.

(a)

The relationship between frequency (f), speed of light (c), and wavelength $\left(\lambda \right)$can be expressed as,

$\mathrm{f}=\frac{c}{\lambda }$…………..(1)

In the above equation (1), substitute the values of c and ${\lambda }_{min}$to get the maximum frequency value, such that

$$\begin{gathered} {\mathrm{f}}_{\max }=\frac{3\times {10}^3\mathrm{m}/\mathrm{s}}{380\times {10}^{-9}\mathrm{m}} \\ =7.89\times {10}^{14}\mathrm{Hz} \end{gathered}$$

In the above equation (1), substitute the values of c and ${\mathrm{\lambda }}_{\max }$to get the maximum frequency value, such that

$$\begin{gathered} {\mathrm{f}}_{\min }=\frac{3\times {10}^{8}m/s}{780\times {10}^{-9}m} \\ =3.84\times {10}^{14}\mathrm{Hz} \\ \mathrm{The}\mathrm{ratio}\mathrm{of}\mathrm{maximum}\mathrm{to}\mathrm{minimum}\mathrm{frequency}\mathrm{can}\mathrm{be}\mathrm{determined}\mathrm{by}, \\ \frac{{\mathrm{f}}_{\max }}{{\mathrm{f}}_{\min }}=\frac{7.89\times {10}^{14}\mathrm{Hz}}{3.84\times {10}^{14}\mathrm{Hz}} \\ =2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{ratio}\mathrm{is}2. \end{gathered}$$

(b)

The ratio of audible highest frequency to audible lowest frequency is calculated as follows:

$$\begin{gathered} \frac{f_{High}}{f_{low}}=\frac{20000\mathrm{Hz}}{20Hz} \\ =1000 \end{gathered}$$

Therefore, the ratio of highest to lowest audible frequency is 1000 .