Answer:
[tex]\displaystyle \lambda = \frac{c}{f}[/tex].
[tex]\lambda \approx 3.33 \times 10^{-8}\; {\rm m}[/tex].
[tex]\displaystyle \lambda_{w} = \frac{c}{n_{w}\, f} \approx 2.56 \times 10^{-8}\; {\rm m}[/tex].
Explanation:
The wavelength [tex]\lambda[/tex] of a wave is the distance travelled in each cycle of the wave. The speed of the wave is the distance travelled in unit time. The frequency [tex]f[/tex] of the wave is the number of cycles (on average) in unit time.
Thus, dividing speed (distance in unit time) by frequency (avg. number of cycles in unit time) would give the distance travelled within each cycle of the wave.
The speed of electromagnetic waves in vacuum is [tex]c[/tex]. Hence, the wavelength of this electromagnetic wave would be:
[tex]\begin{aligned}\lambda &= \frac{c}{f} \\ &\approx \frac{3.00 \times 10^{8}\; {\rm m\cdot s^{-1}}}{9 \times 10^{15}\; {\rm Hz}}\\ &\approx \frac{3.00 \times 10^{8}\; {\rm m\cdot s^{-1}}}{9 \times 10^{15}\; {\rm s^{-1}}} \\ &\approx 3.33 \times 10^{-8}\; {\rm m}\end{aligned}[/tex].
(Note that [tex]1\; {\rm Hz} = 1\; {\rm s^{-1}}[/tex].)
If the speed of light in a particular medium is [tex]v[/tex], the refractive index of that medium would be [tex]n = (c / v)[/tex].
For example, in this question, if the speed of light in water is [tex]v_{w}[/tex], the refractive index of water would be expressed as:
[tex]\displaystyle n_{w} &= \frac{c}{v_{w}}[/tex].
Rearrange this equation to find the speed of light in water, [tex]v_{w}[/tex]:
[tex]\displaystyle v_{w} = \frac{c}{n_{w}}[/tex].
Substitute this expression into the equation for wavelength to find the wavelength of this wave in water:
[tex]\begin{aligned}\lambda_{w} &= \frac{v_{w}}{f} \\ &= \frac{(c / n_{w})}{f} \\ &= \frac{c}{n_{w}\, f} \\ &\approx \frac{3.00 \times 10^{8}\; {\rm m\cdot s^{-1}}}{(1.3)\, (9\times 10^{15}\; {\rm s^{-1}})} \\ &\approx 2.56 \times 10^{-8}\; {\rm m}\end{aligned}[/tex].
