Question

A white light is passed through a soap film of thickness 225 nm and $\mu=1.5$. Which colour of light is seen at reflection?

• A. orange
• B. yellow
• C. Red
• D. Blue

Answer 1

Answer: (D)

Blue

Answer 2

The phenomenon observed is thin-film interference. When white light is incident on a thin film, light reflected from the top surface interferes with light reflected from the bottom surface.

Here's a breakdown of the conditions for constructive interference (bright reflection) in a thin film in air:

1. Reflections and Phase Changes:

   • Light reflecting from the top surface (air to soap film): Since the soap film ($\mu=1.5$) is optically denser than air ($\mu \approx 1$), a phase change of $\pi$ radians (or $\lambda/2$) occurs upon reflection.
   • Light reflecting from the bottom surface (soap film to air): Since air is optically rarer than the soap film, no phase change occurs upon reflection.

2. Optical Path Difference (OPD): The light ray that enters the film travels an extra distance of $2t$ within the film, where $t$ is the thickness. The optical path difference is $2\mu t$, where $\mu$ is the refractive index of the film.

3. Condition for Constructive Interference (Bright Reflection):

   For constructive interference of the reflected light, the total phase difference between the two reflected rays must be an even multiple of $\pi$. Considering the $\pi$ phase change at the top surface reflection and the optical path difference, the condition is:

   $2\mu t = (m + 1/2)\lambda$

   where:

   • $\mu$ is the refractive index of the film.
   • $t$ is the thickness of the film.
   • $\lambda$ is the wavelength of light in air.
   • $m$ is an integer ($0, 1, 2, ...$) representing the order of interference.

Given values:

• Thickness of the soap film, $t = 225$ nm
• Refractive index of the soap film, $\mu = 1.5$

Calculation:

First, calculate the value of $2\mu t$:

$2\mu t = 2 \times 1.5 \times 225 \text{ nm} = 3 \times 225 \text{ nm} = 675 \text{ nm}$

Now, substitute this into the constructive interference equation:

$675 \text{ nm} = (m + 1/2)\lambda$

We need to find the wavelength $\lambda$ that falls within the visible spectrum (approximately 400 nm to 750 nm) for different integer values of $m$:

• For $m = 0$:

  $675 = (0 + 1/2)\lambda$

  $675 = 0.5\lambda$

  $\lambda = \frac{675}{0.5} = 1350 \text{ nm}$ (This is in the infrared region, not visible)

• For $m = 1$:

  $675 = (1 + 1/2)\lambda$

  $675 = 1.5\lambda$

  $\lambda = \frac{675}{1.5} = 450 \text{ nm}$ (This is in the visible spectrum)

• For $m = 2$:

  $675 = (2 + 1/2)\lambda$

  $675 = 2.5\lambda$

  $\lambda = \frac{675}{2.5} = 270 \text{ nm}$ (This is in the ultraviolet region, not visible)

The only visible wavelength that undergoes constructive interference (and is therefore seen) is 450 nm.

The visible spectrum ranges are approximately:

• Violet: 400-450 nm
• Blue: 450-495 nm
• Green: 495-570 nm
• Yellow: 570-590 nm
• Orange: 590-620 nm
• Red: 620-750 nm

A wavelength of 450 nm corresponds to Blue light.