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Consider a soap bubble of thickness ‘d’ and μ = 1.5 Light of λ = 600 nm is incident on the bubble which value of ‘d’ corresponds to destructive interference

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200 nm

300 nm

150 nm

180 nm

answer is C.

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Detailed Solution

$2\mu d{\text{ cosr = n}}\lambda \Rightarrow 2\mu d = n\lambda \$ (For normal difference r = 0^o).

$\therefore d=\frac{n\lambda }{2\mu },n=0,1,2, ....$

$d=0,For \,\,n=0.$ $For\,\, n=1,d= \frac{600}{2\times1.5}= 200\,nm$

$For\,\, n=2,d= \frac{2\times600}{2\times1.5}= 400\,nm$

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