The problem involves thin-film interference, where light reflects off both the top and bottom surfaces of a thin film. The goal is to determine the smallest non-zero film thickness for which destructive interference occurs (resulting in darkness).

Step 1: Understanding the Setup

1. A green light ray of wavelength $\lambda = 450 \, \text{nm}$ strikes a thin film perpendicularly.
2. The refractive index of the film is $n = 1.3$, and it is sandwiched between:
   • Air above ($n_{\text{air}} = 1$),
   • Glass below ($n_{\text{glass}} = 1.5$).

When light reflects off boundaries with different indices, the phase changes must be considered:

• At the air-film interface ($n_{\text{air}} < n_{\text{film}}$): Phase shift of $\pi$ (half-wavelength shift).
• At the film-glass interface ($n_{\text{film}} < n_{\text{glass}}$): Phase shift of $\pi$ again.

Thus, both reflected rays undergo a phase shift of $\pi$.

Step 2: Destructive Interference Condition

For destructive interference in thin films, the optical path difference (OPD) between the two reflected rays must satisfy:

$2 t n_{\text{film}} = m \lambda_{\text{film}},$ where:

• $t$ is the thickness of the film,
• $n_{\text{film}} = 1.3$ is the refractive index,
• $\lambda_{\text{film}}$ is the wavelength of light inside the film,
• $m$ is an integer ($m = 0, 1, 2, \dots$).

The wavelength inside the film is given by:

$\lambda_{\text{film}} = \frac{\lambda_{\text{air}}}{n_{\text{film}}} = \frac{450 \, \text{nm}}{1.3}.$

Step 3: Solving for Minimum Thickness $t$

The smallest non-zero thickness $t$ corresponds to $m = 1$ (first order):

$2 t n_{\text{film}} = \lambda_{\text{film}}.$

Substitute values:

$2 t (1.3) = \frac{450 \, \text{nm}}{1.3}.$

Simplify:

$t = \frac{\frac{450}{1.3}}{2 (1.3)}.$

$t = \frac{450}{2 (1.3)^2}.$

$t = \frac{450}{2 \times 1.69}.$

$t \approx \frac{450}{3.38}.$

$t \approx 133.14 \, \text{nm}.$

Final Answer:

The smallest non-zero thickness of the film for destructive interference is approximately:

$t \approx 133 \, \text{nm}.$

Phase Explanation:

• Both reflections at the air-film and film-glass interfaces undergo a phase shift of $\pi$.
• The condition for destructive interference depends only on the optical path difference, as phase shifts are identical.

Follow-Up Questions:

1. How does the refractive index affect the wavelength of light in a material?
2. Why do phase changes occur at interfaces with different refractive indices?
3. How would the result change if the light were not normal to the surface?
4. What happens if the refractive index of the film increases?
5. Can constructive interference occur in the same thin film? Under what condition?

Tip: For thin films, always consider both phase shifts and path differences to determine interference conditions.