Question

White light, with uniform intensity across the visible wavelength range of 400 nm - 690 nm, is perpendicularly incident on a water film, of index of refraction 1.33 and thickness 300 nm. At what wavelength (in nanometers) is the light reflected by the film brightest to an observer

Answer 1

Answer:

$\lambda=532nm$

Explanation:

From the question we are told that:

Wavelength range of white light  $400 nm - 690 nm$

Index of refraction $n=1.33$

Thickness $T=300*10^{-9}m$

Generally Constructive interference is mathematically given by

$2nd cos\alpha=(m-\frac{1}{2})\lambda\\\\2nd cos\alpha=2nT\\\\Where \alpha=0\textdegree$

Therefore

$For m=12*1.33*300nm=(1-\frac{1}{2})\\\\\lambda=\frac{2*1.33*300nm}{(m-\frac{1}{2})}\\\\\lambda=1596nm\\\\For m=2\\\\2*1.33*300nm=(2-\frac{1}{2})$

$\lambda=532nm$

Therefore the wavelength (in nanometers)  the light reflected by the film brightest to an observer is

$\lambda=532nm$