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 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:

For thin film reflection maxima, use \(2nt = m\lambda\) and adjacent maxima differ by 1 in \(m\).

Answer 1

The Correct Option is B

Step 1: Understand phase change conditions.
At air--soap interface: reflection from rarer to denser, so phase reversal occurs.
At soap--glass interface: reflection from rarer (soap) to denser (glass), so phase reversal occurs again.
Thus, two phase reversals cancel each other.
Step 2: Condition for reflection maxima at normal incidence.
\[ 2nt = m\lambda \]
Step 3: Adjacent maxima correspond to successive integers.
So for \(\lambda_1 = 420\) nm and \(\lambda_2 = 630\) nm:
\[ 2nt = m(420) = (m-1)(630) \]
Step 4: Solve for \(m\).
\[ m\cdot 420 = (m-1)\cdot 630 \]
\[ 420m = 630m - 630 \]
\[ 210m = 630 \Rightarrow m = 3 \]
Step 5: Compute thickness \(t\) using \(m=3\).
\[ 2nt = 3 \times 420 \]
\[ 2(1.4)t = 1260 \Rightarrow 2.8t = 1260 \]
\[ t = \frac{1260}{2.8} = 450 \text{ nm} \]
Final Answer:
\[ \boxed{450 \text{ nm}} \]