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A plastic bubble of thickness 102 nmhas its strongest (m = 1) reflectionmaximum at 604 nm. What is the indexof refraction n for the bubble?(Hint: If you leave the wavelength in nm,the answer will be in nm. No conversionnecessary.)(No unit.)

Answer :

Given:

m = 1

d = 102 nm

λ = 604 nm

n = ?

Apply:

[tex]d=\frac{1}{2}\frac{(m-\frac{1}{2})\lambda}{n}[/tex][tex]n=\frac{(m-1/2)\lambda}{2d}[/tex][tex]n=\frac{(1-1/2)604}{2\text{ \lparen102\rparen}}[/tex]

n= 1.48 nm

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