Question

A thin film of soap solution (\(\mu_s = 1.4\)) 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
• B. 450 nm
• C. 630 nm
• D. 1260 nm

Hint:

When both reflections have phase change of \(\pi\), condition for maxima is \(2\mu t = m\lambda\).

Answer 1

The Correct Option is B

Step 1: Underlying Concept:
For thin film in air, condition for constructive interference: \(2\mu t = m\lambda\). Since film is on glass (\(\mu_g>\mu_s\)), both reflections undergo a phase change of \(\pi\). Step 2: Explanation:
Condition: \(2\mu t = n\lambda_1 = (n-1)\lambda_2\). Substitute: \(n(420) = (n-1)630\). Solve: \(420n = 630n - 630 \Rightarrow 210n = 630 \Rightarrow n = 3\). Find thickness: \[ 2 \times 1.4 \times t = 3 \times 420 \Rightarrow 2.8t = 1260 \Rightarrow t = 450 \text{ nm} \] Step 3: Conclusion:
\[ \boxed{450 \text{ nm}} \]