A soap film of thickness 0.0011mm appears dark when seen by the reflected light of wavelength 580nm. What is the index of refraction of the soap solution, if it is known to be between 1.2 and 1.5?
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Given, $\text{d}=0.0011\times10^{-3}\text{m}$
For minimum reflection of light, $2\mu\text{d}=\text{n}\lambda$
$\Rightarrow\mu=\frac{\text{n}\lambda}{2\text{d}}=\frac{2\text{n}\lambda}{4\text{d}}=\frac{580\times10^{-9}\times2\text{n}}{4\times11\times10^{-7}}$ $=\frac{5.8}{44}(2\text{n})=0.132(2\text{n})$
Given that, $\mu$ has a value in between 1.2 and 1.5.
⇒ When, $\text{n}=5,\mu=0.132\times10=1.32.$
For minimum reflection of light, $2\mu\text{d}=\text{n}\lambda$
$\Rightarrow\mu=\frac{\text{n}\lambda}{2\text{d}}=\frac{2\text{n}\lambda}{4\text{d}}=\frac{580\times10^{-9}\times2\text{n}}{4\times11\times10^{-7}}$ $=\frac{5.8}{44}(2\text{n})=0.132(2\text{n})$
Given that, $\mu$ has a value in between 1.2 and 1.5.
⇒ When, $\text{n}=5,\mu=0.132\times10=1.32.$
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