Question

What is the minimum thickness (in nm) of a soap film ($n = 1.3$) that results in constructive interference in reflected light if the film is illuminated with light whose wavelength in free space is 620 nm?

• A. 100
• B. 120
• C. 160
• D. 240
• E. 180

Hint:

Thin film interference is "inverted" compared to standard interference because of the phase shift at the top surface. Constructive interference occurs at path differences of $\lambda/2, 3\lambda/2$, etc.

Answer 1

The Correct Option is B

Concept:
Reflected light from a thin film undergoes a phase change of $\pi$ at the first surface (air to film). For constructive interference, the path difference $2nt$ must satisfy: \[ 2nt = \left(m + \frac{1}{2}\right)\lambda \quad \text{for } m = 0, 1, 2... \] For minimum thickness ($t$), we take $m = 0$: \[ 2nt = \frac{\lambda}{2} \implies t = \frac{\lambda}{4n} \]

Step 1: Calculate the minimum thickness.
Given $\lambda = 620$ nm and $n = 1.3$: \[ t = \frac{620}{4 \times 1.3} = \frac{620}{5.2} \approx 119.23 \text{ nm} \] Rounding to the nearest whole number, we get 120 nm.