# Oil of refractive index 1.24 makes a film 280 nm thick. Describe what you would expect to see when direct sunlight illuminates this region.

valentin68 | Certified Educator

calendarEducator since 2013

starTop subjects are Science, Math, and History

A ray of white light coming from above will be reflected back by the upper interface of the thin oil film and by the lower interface of the oil thin film. The two reflected light waves will interfere after reflection and a person will see only one color for the light emerging, that is the color for which the interference is constructive.

There are two different cases.

1.

First case is a stand alone oil thin film. (both upper and lower interfaces are with the air)  In this case the light ray that reflects on the upper interface will have a phase shift of +180 degree (or equivalent `lambda/2` ) and the ray that reflects on the lower interface will have zero phase shift. The 180 degree phase shift happens whenever the reflection is made on a medium with higher refractive index. Condition for maximum interference is

`lambda/2 + d*n =k*lambda` ,

where `d =280 nm` is the wavelength, `k` is an integer and `n=1.24` is the refractive index of oil.

Taking `k=1` one has

`d*n = lambda/2`  or equivalent

`lambda=2*d*n =2*280*1.24 =694.4 nm` , red color

Taking `k=2` one has `d*n =3*lambda/2` or equivalent

`lambda=2/3*d*n =2/3*280*1.24 =231.44 nm` , ultra violet (not visible)

Hence in this case the person looking at the oil film will see only the red color.

2.

Second case is an oil film standing on water (`n_(water)= 1.33)` . In this case both reflected light rays will have the same phase. (none will lose 180 degree at reflection).
The condition for constructive interference is

`d*n =k*lambda `

For `k=1` we obtain `lambda = 280*1.24 =347.2 nm` , Violet color

For `k=2` we obtain `lambda=(d*n)/2 =280*1.24/2 =173.6 nm` , ultraviolet not visible

In this case the person looking at the oil film will see only the violet color.

check Approved by eNotes Editorial