Geometrical problem related to thin film interference

[khaos89]

Look at the picture below, I have to prove that the optical path length difference is

[itex] \Delta=n(BC+CD)-BE=2nd\cos(r)[/itex]

[PLAIN]http://img200.imageshack.us/img200/2271/schermata082455775alle1.th.png

The problem is just how to get [itex]2nd\cos(r)[/itex]

I actually don't have any idea :\

I have tried to work with trigonometry but no luck yet...

(I am posting it here because it's not related to course work, i am just trying to understand how it works when we don't have perpendicular incidence)

 

[khaos89]

Sorry, here we go with the pic:

[PLAIN]http://img193.imageshack.us/img193/2271/schermata082455775alle1.png

 

[GenePeer]

http://en.wikipedia.org/wiki/Thin-film_interference

Can't believe I'd forgotten it already! Half a year doing nothing is bad.

 

[khaos89]

Thanks a lot :)

 

[CellCoree]

I can provide some insight into this geometrical problem related to thin film interference. Firstly, let's define some terms to help understand the problem better. The optical path length difference (OPLD) is the difference in distance that light travels between two paths. In this case, it is the difference in distance traveled by the light between the upper and lower paths in the thin film. The index of refraction (n) is the ratio of the speed of light in a vacuum to the speed of light in a given medium. The angle (r) is the angle of incidence, which is the angle at which the light enters the thin film.

Now, let's look at the diagram provided. The OPLD is equal to the difference in distance traveled by light in the upper and lower paths. In the upper path, the light travels from point B to point C, then from point C to point D. This distance can be represented as n(BC+CD), where n is the index of refraction and BC+CD is the total distance traveled in the upper path.

In the lower path, the light travels from point B to point E, then from point E to point D. This distance can be represented as BE+ED. However, since BE and ED are parallel lines, they are equal in length. Therefore, BE=ED=d, where d is the thickness of the thin film.

Now, we can substitute this into the equation for the lower path: BE+ED=2d. Therefore, the OPLD is equal to n(BC+CD)-2d.

Next, we need to consider the angle of incidence (r). In the case of perpendicular incidence, r=0 and the OPLD simplifies to n(BC+CD)-BE. However, in this problem, the angle of incidence is not perpendicular, so we need to factor in the cosine of the angle of incidence (cos(r)).

Using trigonometry, we can see that the length of BE can be represented as dcos(r). Therefore, the OPLD can be rewritten as n(BC+CD)-dcos(r). And since we already know that BE=ED=d, we can combine these terms to get the final equation: OPLD=2ndcos(r).

In conclusion, we have successfully proved that the OPLD is equal to 2ndcos(r) in this geometrical problem related to thin