Answer:
[tex]\lambda=532nm[/tex]
Explanation:
From the question we are told that:
Wavelength range of white light [tex]400 nm - 690 nm[/tex]
Index of refraction [tex]n=1.33[/tex]
Thickness [tex]T=300*10^{-9}m[/tex]
Generally Constructive interference is mathematically given by
[tex]2nd cos\alpha=(m-\frac{1}{2})\lambda\\\\2nd cos\alpha=2nT\\\\Where \alpha=0\textdegree[/tex]
Therefore
[tex]For m=12*1.33*300nm=(1-\frac{1}{2})\\\\\lambda=\frac{2*1.33*300nm}{(m-\frac{1}{2})}\\\\\lambda=1596nm\\\\For m=2\\\\2*1.33*300nm=(2-\frac{1}{2})\\\\[/tex]
[tex]\lambda=532nm[/tex]
Therefore the wavelength (in nanometers) the light reflected by the film brightest to an observer is
[tex]\lambda=532nm[/tex]
