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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

Sagot :

Okpalawalter8idn

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]

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