Show that when light passes from air to water, its wavelength decreases to\({\rm{0}}{\rm{.750}}\)times its original value.

The wavelength is defined as the distance between two identical locations, i.e., minima or maxima in a wave pattern.

The frequency of the wave remains constant.

The speed of an electromagnetic wave can be defined by the formula,

$\frac{\mathrm{c}}{\mathrm{\lambda }}=\mathrm{v}$

Where ${}^v$and ${}^{\lambda }$are the frequency and wavelength of the light, respectively.

The visible light spectrum ranges from the wavelengths ${}^{380\mathrm{nm}}$to ${}^{760\mathrm{nm}}$

The wavelength of an electromagnetic wave in a medium can be defined as,

$\mathrm{\lambda }=\frac{\mathrm{\lambda }}{\mathrm{n}}$

where${}^{\lambda }$is the wavelength of the electromagnetic wave in a vacuum and${}^{\mathrm{n}}$is the refraction index of the specified medium.

The light travels from the air to the water and the water's refractive index is,

$n_{water}=1.333$

In a medium, the wavelength of an electromagnetic wave is given by,

$$\begin{gathered} {\lambda }_3=\frac{\lambda }{n_{water}} \\ =\frac{\lambda }{1.333} \\ =0.750\lambda \end{gathered}$$

Therefore, the wavelength decreases to${}^{0.750\lambda }$ as compared to its original value.