As per the question,
The wavelength of the light (λ)=460nm(\lambda)=460nm(λ)=460nm
n =320 dark fringe
Let the thickness be e,
e=mλ2e=\frac{m\lambda }{2}e=2mλ
⇒e=320×460×10−9m2\Rightarrow e=\frac{320\times 460\times 10^{-9}m}{2}⇒e=2320×460×10−9m
⇒e=73.6×10−6m\Rightarrow e=73.6\times 10^{-6}m⇒e=73.6×10−6m
⇒e=73.6μm\Rightarrow e =73.6\mu m⇒e=73.6μm
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