Question

A thin film of soap solution $(n = 1.4)$ lies on the top of a glass plate $(n = 1.5)$. When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths $420$ and $630\, nm$. The minimum thickness of the soap solution is

• A. 420 nm
• B. 450 nm
• C. 630 nm
• D. 1260 nm

Answer 1

Answer: (B)

$n_{1} \lambda_{1}=n_{2} \lambda_{2}$
$\therefore n_{1} \times 420 =n_{2} \times 630$
or $ 2 n_{1}=3 n_{2}$
If $ n_{2}=2 $ then $ n_{1}=3$
Therefore, thickness of soap solution is given by
or $\mu_{1} t =n_{1} \frac{\lambda_{1}}{2} $
$t =\frac{3 \times 420}{1.4 \times 2}=450\, nm$