Problem 18 - Physics

Question

A uniform layer of water \((n=1.33)\) lies on a glass plate \((n=1.52)\). Light shines perpendicularly on the layer. Because of constructive interference, the layer looks maximally bright when the wavelength of the light is \(432 \mathrm{~nm}\) in vacuum and also when it is \(648 \mathrm{~nm}\) in vacuum. (a) Obtain the minimum thickness of the film. (b) Assuming that the film has the minimum thickness and that the visible spectrum extends from 380 to \(750 \mathrm{~nm}\), determine the visible wavelength(s) (in vacuum) for which the film appears completely dark.

Step-by-Step Solution

Answer

(a) Minimum thickness is approximately 162.41 nm. (b) Find visible spectrum dark wavelengths using destructive interference conditions.

Step 1: Understand the Problem

We have a thin film of water on glass, with different indices of refraction. Light shines perpendicularly to the film, and due to constructive interference, certain wavelengths appear bright. We're tasked to find the minimum thickness of this film for these bright wavelengths, and to determine when the film appears dark within a given spectrum.

Step 2: Determine Constructive Interference Condition

For constructive interference in a thin film, the condition is given by \[ 2nt = m\lambda, \]where \( n \) is the index of refraction of the film, \( t \) is the thickness of the film, \( m \) is the order of the interference, and \( \lambda \) is the wavelength in vacuum. Here, the two wavelengths given are 432 nm and 648 nm.

Step 3: Calculate Film Thickness from Given Wavelengths

Use the constructive interference formula \[ 2nt = m\lambda. \]For the wavelengths, consider them as successive orders due to the minimal thickness assumption. Assume for \( \lambda_1 = 432 \text{ nm} \), \( m = 1 \) and for \( \lambda_2 = 648 \text{ nm} \), \( m = 2 \). Use these to find \( t \). For \( n = 1.33 \), \[ 2 \times 1.33 \times t = 432\text{ nm}. \]Solve for \( t \) to get \[ t = \frac{432}{2 \times 1.33} \approx 162.41\text{ nm}. \]

Step 4: Determine Dark Wavelengths, Applying Destructive Interference

For destructive interference, the condition is given by \[ 2nt = (m + 0.5)\lambda. \]With the film thickness as obtained and for the visible spectrum (380 to 750 nm), evaluate wavelengths for which destructive interference occurs.

Step 5: Calculate for Destructive Interference Condition

For a given value of \( t \), substitute into the condition \[ 2 \times 1.33 \times 162.41 = (m+0.5) \lambda \]to get possible wavelengths. You need to make different assumptions for \( m \) and find such wavelengths within the visible range.

Key Concepts

Constructive Interference

Constructive interference occurs when waves combine to make a new wave with a larger amplitude, resulting in a bright appearance. In thin film interference, constructive interference happens when light waves reflected off the top and bottom surfaces of a film are exactly in phase. In technical terms, this means their peaks and troughs line up perfectly.
This phenomenon can be described mathematically by the equation:\[ 2nt = m\lambda, \]where:

• \( n \) is the refractive index of the film,
• \( t \) is the thickness of the film,
• \( m \) is the interference order, indicating the number of times the waves "fit" into the film,
• \( \lambda \) is the wavelength of light in a vacuum.

To achieve constructive interference, the wave path difference must be an integer multiple of the wavelength. In simpler terms, the light that travels through the film must return in a way that enhances the total light intensity instead of reducing it.

Destructive Interference

Destructive interference happens when waves combine to cancel each other out, making the resulting wave smaller or even eliminating it entirely. In the context of thin films, when light waves reflecting off the surfaces are out of phase, their peaks and troughs do not align, leading to a reduction in light intensity or darkness.
The formula describing destructive interference is:\[ 2nt = (m + 0.5)\lambda, \]highlighting that the reflected light has path difference corresponding to half a wavelength difference from complete cycles. This means the peaks of one wave coincide with the troughs of another.
Here's a breakdown of the variables:

• \( n \) is the refractive index of the film,
• \( t \) represents the film thickness,
• \( m \) is the order of interference, representing full wavelengths fitting into the film path difference,
• \( \lambda \) is the wavelength in a vacuum.

With thin films, various "dark spots" occur at different thicknesses and wavelengths depending on these factors, defining when minimal or no reflection happens in the visible spectrum.

Wavelength

A wavelength is the distance between successive peaks of a wave, usually measured in nanometers. Light, a form of electromagnetic wave, has its own characteristic wavelengths, each corresponding to different colors in the visible spectrum.
The visible spectrum covers wavelengths approximately from 380 nm (violet) to 750 nm (red). The specified wavelengths often dictate the behavior of interference effects in thin films:

• Shorter wavelengths (near violet end) correspond to higher energy light, and often shorter path differences create interference effects.
• Longer wavelengths (near red end) require different conditions for interference to occur due to their longer path lengths and different energy levels.

Understanding how different wavelengths interact with materials' thickness and refractive index is crucial for solving problems in optics, like calculating the minimum thickness of thin films or determining conditions for dark or bright appearances. These factors influence how light is reinforced (bright) or reduced (dark) at certain thicknesses and incidence angles in materials.