What is the Formula for Thin Film Interference?

AI Thread Summary
The discussion focuses on the formula for thin film interference, specifically addressing the calculations for destructive interference. A participant calculated the film thickness by dividing the wavelength of 425 nm by 4, resulting in 106 nm, and sought confirmation of this method. The conversation emphasizes the importance of understanding phase shifts during reflection, noting that a phase shift of pi occurs when light reflects off a medium of higher optical density. The provided formulas for constructive and destructive interference highlight the relationship between film thickness, wavelength, and phase shifts. Overall, the thread clarifies the principles behind calculating thin film interference and the significance of phase shifts in these calculations.
Precursor
Messages
219
Reaction score
0
Homework Statement

[PLAIN]http://img403.imageshack.us/img403/2696/33095583.jpg

The attempt at a solution

I simply took 425 nm and divided it by 4 by using the destructive reflection from thin film. I got 106 nm as my answer, which is c. Is this correct?
 
Last edited by a moderator:
Physics news on Phys.org
Precursor said:
I simply took 425 nm and divided it by 4 by using the destructive reflection from thin film.
What's your reasoning behind this?
 
remember your rules of reflection. when an electromagnetic wave is in incidence of a medium of higher optical density the light will have a phase shift of pi. If an electromagnetic wave is incidence on a less optically dense medium there is no shift.

contructive 2mpi = (4(pi)ntcos(x))/lambda -k + b

destructive (2q+1)pi = (4(pi)ntcos(x))/lambda -k + b

where is the phase shift due to the intial and main medium and b is the phase shift due to the main medium and final. x is the angle of incidence which would be 0 in your case. t is film thickness of main medium. n is the index. lambda is wavelength.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How Does Thin Film Interference Affect Light Reflection?
Replies
1
Views
2K
Finding the Minimum Thickness for Destructive Interference in Thin Films
Replies
7
Views
6K
Thin film interference and external / internal reflection
Replies
1
Views
2K
What is the equation for constructive interference in thin film interference?
Replies
3
Views
2K
Formula for maximum interference for reflected light (thin - film)
Replies
3
Views
2K
Thin film constructive interference
Replies
3
Views
4K
The maximum intensity for light transmitted through a thin film
Replies
2
Views
3K
What Visible Wavelengths Cause Maximum Constructive Interference in a Thin Film?
Replies
1
Views
3K
Interference in Thin Films: Grade 12 Optics Review
Replies
4
Views
6K
Thin-Film Interference (2 Wavelengths)
Replies
14
Views
3K

Hot Threads

Recent Insights

Back
Top