Question

There is soap film of thickness 1.5 $\mu m$ of refractive index 4/3. Considering visible light wavelengths $\lambda \in (3800 \mathring{A} \text{ to } 7600 \mathring{A})$, then (consider normal incidence)

• A. Wavelengths which are strongly reflected are 6 in number
• B. Wavelengths missing in reflecting spectrum are 5 in number
• C. Wavelengths which are strongly refracted are 5
• D. Wavelengths which are relatively weakly refracted are 5

Answer 1

Answer: (A), (B), (C)

A, B, and C are correct.

Answer 2

The phenomenon is thin-film interference. For normal incidence, the optical path difference (OPD) between the two reflected rays is $2\mu t$.

Given: Thickness of the film, $t = 1.5 \, \mu m = 1.5 \times 10^{-6} \, m$ Refractive index of the film, $\mu = 4/3$ Visible light range: $\lambda \in (3800 \, \mathring{A} \text{ to } 7600 \, \mathring{A})$, which is $380 \, nm \text{ to } 760 \, nm$.

Calculate $2\mu t$: $2\mu t = 2 \times \frac{4}{3} \times (1.5 \times 10^{-6} \, m) = \frac{8}{3} \times 1.5 \times 10^{-6} \, m = 8 \times 0.5 \times 10^{-6} \, m = 4 \times 10^{-6} \, m = 4000 \, nm$.

Phase Shifts:

1. Reflection at the top surface (air to film): Since $\mu_{film} > \mu_{air}$, there is a phase change of $\pi$.
2. Reflection at the bottom surface (film to air): Since $\mu_{air} < \mu_{film}$, there is no phase change.

1. Reflected Light:

• Constructive Interference (Bright Reflection): $2\mu t = (m + 1/2)\lambda$ $\lambda = \frac{2\mu t}{m + 1/2} = \frac{4000}{m + 0.5} \, nm$. For visible light ($380 \le \lambda \le 760$): $m \ge 4.76$ and $m \le 10.03$. Possible integer values for $m$ are $5, 6, 7, 8, 9, 10$. There are 6 wavelengths for constructive interference (strongly reflected). Option A is correct.

• Destructive Interference (Dark Reflection / Missing in Reflection): $2\mu t = n\lambda$ $\lambda = \frac{2\mu t}{n} = \frac{4000}{n} \, nm$. For visible light ($380 \le \lambda \le 760$): $n \ge 5.26$ and $n \le 10.53$. Possible integer values for $n$ are $6, 7, 8, 9, 10$. There are 5 wavelengths for destructive interference (missing in reflection). Option B is correct.

2. Transmitted Light: OPD = $2\mu t$. Total phase difference = $\frac{2\pi}{\lambda}(2\mu t)$.

• Constructive Interference (Bright Transmission / Strongly Refracted): $\frac{2\pi}{\lambda}(2\mu t) = 2n\pi \implies 2\mu t = n\lambda$ $\lambda = \frac{4000}{n} \, nm$. As calculated for destructive reflection, possible integer values for $n$ are $6, 7, 8, 9, 10$. There are 5 wavelengths for constructive interference (strongly refracted). Option C is correct.

• Destructive Interference (Dark Transmission / Weakly Refracted): $\frac{2\pi}{\lambda}(2\mu t) = (2n+1)\pi \implies 2\mu t = (m + 1/2)\lambda$ $\lambda = \frac{4000}{m + 0.5} \, nm$. As calculated for constructive reflection, possible integer values for $m$ are $5, 6, 7, 8, 9, 10$. There are 6 wavelengths for destructive interference (weakly refracted). Option D is incorrect.