Question

(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.

Short answer

1. The ratio of highest to lowest visual frequency is 2 .
2. The ratio of highest to lowest audible frequency is 1000 .

Step-by-step solution

Define electromagnetic waves

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

Evaluating the ratio of highest and lowest visible frequency

(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

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

Evaluating the ratio of highest to the lowest audible frequency

(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 .