Explanation:
: Given, \(\mu_{\mathrm{W}}=1.33\), thickness \((\mathrm{t})=320 \mathrm{~nm}\) Solving for maxima-
\(\lambda=\frac{2 \mu \mathrm{t}}{\left(\mathrm{n}+\frac{1}{2}\right)}=\frac{2 \times 1.33 \times 320 \mathrm{~nm}}{\left(\mathrm{n}+\frac{1}{2}\right)}=\frac{851 \mathrm{~nm}}{\left(\mathrm{n}+\frac{1}{2}\right)}\)
for, \(\mathrm{n}=0, \lambda=1700 \mathrm{~nm}\) (infrared region)
for, \(\mathrm{n}=1, \lambda=567 \mathrm{~nm}\) (visible region)
for, \(\mathrm{n}=2, \lambda=340 \mathrm{~nm}\) (ultraviolet)
So, for \(\mathrm{n}=1\), the wavelength at which light seen by the observer is brightest, \(\lambda=567 \mathrm{~nm}\)