Problem 19 - Master Resource Book in JEE Main Physics

Question

A light of wavelength 5890 A falls normally on a thin air film. The minimum thickness of the film such that the film appears dark in reflected light is (a) \(2.945 \times 10^{-7} \mathrm{~m}\) (b) \(3.945 \times 10^{-7} \mathrm{~m}\) (c) \(4.95 \times 10^{-7} \mathrm{~m}\) (d) \(1.945 \times 10^{-7} \mathrm{~m}\)

Step-by-Step Solution

Answer

The minimum thickness is closest to 1.945 × 10⁻⁷ m, option (d).

Step 1: Understand the Concept of Thin Film Interference

When light is incident on a thin film, such as an air film, interference can occur between light waves reflected from the top and bottom surfaces of the film. For a dark fringe to appear in reflected light, the path difference between the two reflected light waves should be an odd multiple of half the wavelength in the film.

Step 2: Express the Condition for Destructive Interference

The condition for minimum thickness such that the film appears dark is given by the equation \[2nt = (m + \frac{1}{2})\lambda,\] where \(n\) is the refractive index of the film, \(t\) is the minimum thickness, \(m\) is an integer (order of interference, starting from 0), and \(\lambda\) is the wavelength of light in vacuum.

Step 3: Adapt the Equation for Air Film with n=1

Since the film is air, its refractive index \(n=1\). Substituting \(n=1\) and \(m=0\) for the first dark fringe, the equation becomes \[2 \cdot 1 \cdot t = \left(0 + \frac{1}{2}\right) \lambda,\] simplifying to \[t = \frac{\lambda}{4}.\]

Step 4: Convert Wavelength to Meters

The wavelength of light is given as 5890 Angstroms. First, convert it to meters: \(1\text{ A} = 10^{-10}\text{ m}\), so \(\lambda = 5890 \times 10^{-10} \text{ m} = 5.89 \times 10^{-7} \text{ m}.\)

Step 5: Calculate Minimum Thickness

Substitute \(\lambda = 5.89 \times 10^{-7} \text{ m}\) into the formula \(t = \frac{\lambda}{4}\) to find \[t = \frac{5.89 \times 10^{-7} \text{ m}}{4} = 1.4725 \times 10^{-7} \text{ m}.\]

Step 6: Identify Closest Answer

Review the provided answer choices: (a) \(2.945 \times 10^{-7} \text{ m}\), (b) \(3.945 \times 10^{-7} \text{ m}\), (c) \(4.95 \times 10^{-7} \text{ m}\), and (d) \(1.945 \times 10^{-7} \text{ m}\). The closest to the calculated value \(1.4725 \times 10^{-7} \text{ m}\) is \(1.945 \times 10^{-7} \text{ m}\), making choice (d) the correct answer.

Key Concepts

Wavelength of Light

The wavelength of light is a critical concept when discussing thin film interference, which describes how light behaves when it passes through a medium. Light is an electromagnetic wave, and its wavelength is the distance over which the wave's shape repeats, often measured in nanometers (nm) or Angstroms (A). In the context of this exercise, the wavelength given is 5890 Angstroms.

• 1 Angstrom (A) is equal to 10^-10 meters.
• This means 5890 Angstroms converts to 5.89 x 10^-7 meters.

Understanding the wavelength is important because it affects how light interferes within a film. It determines the appearance of colors or darkness when looking at the film in specific lighting conditions, as seen in examples like soap bubbles or oil spills on water.

Destructive Interference

Destructive interference is when two or more waves combine in such a way that they cancel each other out.This results in a reduction of the wave's amplitude, meaning the waves effectively "destroy" each other's effects.For a thin film, such as the air film in the exercise:

• The light waves reflected from the top and bottom surfaces interfere with each other.
• Destructive interference occurs at specific thicknesses, creating dark fringes.

The mathematical condition for destructive interference in thin films is given by the equation:\[2nt = (m + \frac{1}{2})\lambda\]where "n" is the refractive index, "t" is the thickness, "\(m\)" is an integer, and "\(\lambda\)" is the wavelength of light.

Refractive Index

Refractive index (n) is a measure of how much light slows down when it enters a medium compared to its velocity in vacuum. It is defined as the ratio of the speed of light in a vacuum to its speed in that medium.

• For air, the refractive index is approximately 1.
• This means light in air moves at nearly the same speed as it does in a vacuum.
• In our exercise, because the film is an air film, we use n=1 in calculations.

The refractive index can be more significant in other mediums, such as water or glass, where light slows more dramatically. Knowing the correct refractive index is key for calculations in interference problems.

Path Difference

Path difference refers to the difference in the travel paths of two light waves before they combine.In thin film interference, it is the difference in distances that the waves travel between being reflected from the two surfaces of the film.

• A larger path difference can lead to a full wave cycle and constructive interference.
• Specific path differences lead to destructive interference, creating dark spots.

The path difference helps determine the condition for interference: for destructive interference, the path difference should be an odd multiple of half-wavelength in the medium.When the refractive index is considered, the formula becomes:\[2nt = (m + \frac{1}{2})\lambda\] This is used to calculate the thickness "t" of a film at which the dark fringes are observed.