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

• A

  420 nm
• Correct answer
  B

  450 nm
• C

  630 nm
• D

  1260 nm

Solution

For the reflection at the air-soap solution interface, the phase difference is $\pi$
For reflection at the interface of soap solution to the glass also there will be a phase difference of $\pi$. $\therefore$ The condition for maximum intensity $=2 \mu t=n \lambda$ ...(i) $\begin{array}{rlrl} & \text { For } n , & n \lambda_1 & =(n-1) \lambda_2 \\ & & n(420) & =(n-1) 630 \\ & & n(630-420) & =630 \\ \therefore & n(210) & =630 \\ \Rightarrow & & n & =3\end{array}$ This is the maximum order where they coincide, From Eq. (i), This is the maximum order where they coincide, From Eq. (i), $\begin{array}{ll} \therefore & 2 \times 1.4 \times t=3 \times 420 \\ \Rightarrow & t=\frac{2 \times 420}{2 \times 1.40}=450 \mathrm{~nm} \end{array}$

Hence, the correct answer is option: 'B'.