Q144 / 160 · TS EAMCET 2018 (05 May Shift 2) · Physics
Calculate the minimum thickness of a soap film ( $n=1.33)$ that results in constructive interference in reflected light, if the film is illuminated with light whose wavelength in free space is $532 \mathrm{~nm}$.
- A$113 \mathrm{~nm}$
- Correct answerB$100 \mathrm{~nm}$
- C$200 \mathrm{~nm}$
- D$226 \mathrm{~nm}$
Solution
For constructive interference through a thin film in reflected light, thickness of film must be
$$
t=\left(m+\frac{1}{2}\right) \frac{\lambda}{2 n_2}
$$
where, $m=0,1,2, \ldots$
and $n_2=$ refractive index of medium (soap film)
With given values, we get
$$
\begin{aligned}
t & =\frac{\lambda}{4 n_2} \quad\left(t_{\min } \text { occurs with } m=0\right) \\
& =\frac{532 \times 10^{-9}}{4 \times 1.33}=100 \mathrm{~nm}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.