Courses for Kids
Free study material
Offline Centres
Store Icon

A soap film in the air has a thickness of 175 nm. If the index of refraction of the soap film is $1.35$, what colour of light will be transmitted through? (i.e., What colour does the bubble appear to be from the other side?

Last updated date: 18th Jun 2024
Total views: 53.1k
Views today: 0.53k
53.1k+ views
Hint: Thin-film interference is the phenomenon that is a result of a light wave being reflected off two surfaces that are at a distance comparable to its wavelength. When light waves that reflect off the top and bottom surfaces interfere with one another we see different coloured patterns. During this, the light reaches the boundary between two media and part of it gets reflected and some part gets transmitted.
We need to calculate the wavelength transmitted and match it with the corresponding wavelength in the visible spectrum region.

Complete Step by Step Solution: The bright colours seen in an oil slick floating on water or in a sunlit soap bubble are caused by interference. The brightest colours are those that interfere constructively. This interference is between light reflected from different surfaces of a thin film; thus, the effect is known as thin film interference.
Incident light is only partially reflected from the top surface of the film. The remainder enters the film and is itself partially reflected from the bottom surface. Part of the light reflected from the bottom surface can emerge from the top of the film and interfere with light reflected from the top. Since the ray that enters the film travels a greater distance, it may be in or out of phase with the ray reflected from the top.
When light reflects from a medium having an index of refraction greater than that of the medium in which it is travelling, a ${180^ \circ }$ phase change (or a $\lambda /2$ shift) occurs.
It has been given that a soap film in the air has a thickness of 175 nm and the index of refraction of the soap film is $1.35$.
The light transmitted through the soap bubble occurs for the same condition as for minimum reflection and the transmitted light will be in phase with the incident light. The light reflected from the internal surface will also be in phase with the incident light.
In order for constructive interference to occur in the transmission, the path length difference must be $l$, so that the film thickness must be $\lambda /2$.
Thus, the wavelength of the light in the film is double the thickness of the film.
And as per the given criteria, ${\mu _{film}} = 1.35$ and ${t_{film}} = 174nm$,
${\lambda _{film}} = 2t$
$ \Rightarrow {\lambda _{film}} = 2 \times 175nm = 350nm$
By the laws of refraction, it is known that, ${\mu _{film}} = \dfrac{{{\lambda _{air}}}}{{{\lambda _{film}}}}$
${\lambda _{air}} = {\mu _{film}} \times {\lambda _{film}}$
$ \Rightarrow {\lambda _{air}} = 1.35 \times 350nm = 472.5nm$
Light with wavelength 473nm will be transmitted through and it is the wavelength of blue colour.
The reflected light is weak in the red region of the spectrum and strong in the blue-violet region. The soap film will, therefore, possess a pronounced blue colour.

Note: The bubbles are darkest where they are thinnest. Furthermore, if a soap bubble is observed carefully, it is seen that it is dark at the point where it breaks.
Thus, when the film is very thin, the path length difference between the two rays is negligible, they are exactly out of phase, and destructive interference will occur at all wavelengths and so the soap bubble will be dark here.