Destructive Interference of Light Waves in Bubble Film (165 nm)

Click For Summary
SUMMARY

The discussion focuses on the conditions for destructive interference of light waves in a bubble film with a thickness of 165 nm and a refractive index of 1.3. The key formula for determining the wavelengths that will experience destructive interference is given by the equation |(1/λ1) - (1/λ2)| = (2k + 1) / (2nd), where d is the film thickness and n is the refractive index. Specific wavelengths mentioned include 430 nm, 650 nm, 860 nm, and 285 nm. The user successfully grasps the concept after receiving guidance on applying the formula.

PREREQUISITES
  • Understanding of wave interference principles
  • Knowledge of refractive index and its implications
  • Familiarity with the concept of wavelength in light waves
  • Basic algebra for manipulating equations
NEXT STEPS
  • Explore the principles of wave optics and interference patterns
  • Learn about the applications of thin film interference in real-world scenarios
  • Study the effects of varying refractive indices on light behavior
  • Investigate the mathematical derivation of interference conditions for different film thicknesses
USEFUL FOR

Students and educators in physics, optical engineers, and anyone interested in the principles of light wave interference and its applications in thin films.

stickplot
Messages
83
Reaction score
0
Ok so I've been thinking this problem over and over and I understand the problem and its concept but I just don't know where to start and how to do this, someone please guide me through this

Light waves of which wavelength will destructively interfere due to the thickness of the bubble film? The bubble film thickness is 165 nm, with an index of refraction of n = 1.3.

It also contains this picture just showing the different light waves

a= 430 nm
b= 650 nm
c= 860 nm
d= 285 nm
 

Attachments

  • light.gif
    light.gif
    9 KB · Views: 457
Physics news on Phys.org
Destructive interference occurs when the phase difference between the two waves is 180 degrees i.e. \pi

That is, if the two wavelengths are \lambda_1\;\;\;,\;\;\; \lambda_2, then the condition for destructive interference is:

\left|\frac{2\pi}{\lambda_1} nd - \frac{2\pi}{\lambda_2}nd\right|=(2k+1)\pi

Where d is the size of the soap film, and n is the refractive index. k is a positive integer.
so from here we have:

\left|\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right|=\frac{2k+1}{2nd}

See for which wavelengths this is satisfied, and you are done...
 
o ok thank you very much i understand it now
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
Replies
7
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
2
Views
4K
Replies
4
Views
3K