Question

An oil film in air of thickness 255 nm is illuminated by white light at normal incidence. As a consequence of interference, which colour will be predominantly visible in the reflected light ?
Given the refractive index of oil = 1.47

• A. Red (~ 650 nm)
• B. Blue (~ 450 nm)
• C. Green (~ 500 nm)
• D. Yellow (~560 nm)

Answer 1

The Correct Option is C

The problem involves determining the color that will be predominantly visible in the reflected light when white light is shone on an oil film of certain thickness, due to the phenomenon of thin-film interference.

Thin-film interference occurs because light waves reflected from the upper and lower boundaries of a film interfere with one another. Depending on the thickness of the film and the wavelength of light, this interference can be constructive or destructive.

In this case, the oil film has a thickness \( t = 255 \) nm and a refractive index \( n = 1.47 \). The condition for constructive interference (a bright fringe) in this case (normal incidence) is given by:

\(2nt = m\lambda\) 

Here, \( \lambda \) is the wavelength of light in vacuum (or air) and \( m \) is the order of interference.

To find the wavelength that will constructively interfere, we solve for \( \lambda\):

\(\lambda = \frac{2nt}{m}\)

Assuming the first order of interference (\( m = 1 \)) for simplicity, the wavelength of visible light that will constructively interfere is:

\(\lambda = \frac{2 \times 1.47 \times 255 \, \text{nm}}{1} = 749.7 \, \text{nm}\)

The wavelength computed (749.7 nm) falls outside the visible spectrum. Hence, we consider the effect of constructive interference at other integer multiples.

Reevaluating for \( m = 2 \) gives:

\(\lambda = \frac{2 \times 1.47 \times 255 \, \text{nm}}{2} = 374.85 \, \text{nm}\)

This wavelength is too low (ultraviolet region). Hence, consider \( m = 3 \).

\(\lambda = \frac{2 \times 1.47 \times 255 \, \text{nm}}{3} = 249.9 \, \text{nm}\)

Again, not in the visible spectrum. Now, let's try \( m = 4 \):

\(\lambda = \frac{2 \times 1.47 \times 255 \, \text{nm}}{4} = 562.275 \, \text{nm}\)

This is within the visible range, close to the green spectrum (~560 nm). Therefore, the predominant color seen would be close to green.

Thus, the color predominantly visible due to thin-film interference in the reflected light is Green (~500 nm).