KCET · Physics · Wave Optics
What is the minimum thickness of a thin film required for constructive interference in the reflected light from it?
Given, the refractive index of the film \(=1.5\), wavelength of the light incident on the film \(=600 \mathrm{~nm}\).
- A \(100 \mathrm{~nm}\)
- B \(300 \mathrm{~nm}\)
- C \(50 \mathrm{~nm}\)
- D \(200 \mathrm{~nm}\)
Answer & Solution
Correct Answer
(A) \(100 \mathrm{~nm}\)
Step-by-step Solution
Detailed explanation
Condition for constructive interference is
\[
2 \mu \mathrm{t}=[2 \mathrm{n}+1] \frac{\lambda}{2}
\]
where, \(\quad \mathrm{n}=0,1,2,3 \ldots\)
For minimum thickness, \(\mathrm{n}=0\)
\[
\begin{gathered}
2 \mu \mathrm{t}=\frac{\lambda}{2} \\
\Rightarrow \quad \mathrm{t}=\frac{\lambda}{4 \mu}=\frac{600 \times 10^{-9}}{4 \times 1.5} \\
=100 \mathrm{~nm}
\end{gathered}
\]
\[
2 \mu \mathrm{t}=[2 \mathrm{n}+1] \frac{\lambda}{2}
\]
where, \(\quad \mathrm{n}=0,1,2,3 \ldots\)
For minimum thickness, \(\mathrm{n}=0\)
\[
\begin{gathered}
2 \mu \mathrm{t}=\frac{\lambda}{2} \\
\Rightarrow \quad \mathrm{t}=\frac{\lambda}{4 \mu}=\frac{600 \times 10^{-9}}{4 \times 1.5} \\
=100 \mathrm{~nm}
\end{gathered}
\]
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