Double Slit Question with Thin Lens concept

Click For Summary
SUMMARY

The discussion centers on a double slit experiment involving a light source of 540nm and a slit separation of 4mm, where a thin glass plate with a refractive index of 1.56 is placed in front of one slit. The objective is to determine the minimum thickness of the glass that causes each bright fringe to shift to the position of the adjacent bright fringe. The relevant equations include 2nt = ±mλ for bright fringes and 2nt = ±(m + ½)λ for dark fringes, leading to the conclusion that the minimum thickness can be calculated using the formula t = λ/(4n), resulting in a thickness of 8.65×10^-8 m.

PREREQUISITES
  • Understanding of the double slit experiment and interference patterns.
  • Familiarity with the concept of optical path difference.
  • Knowledge of the refractive index and its impact on light propagation.
  • Basic proficiency in using equations related to wave optics.
NEXT STEPS
  • Research the implications of optical path difference in interference patterns.
  • Learn about the effects of varying refractive indices on light behavior in thin films.
  • Explore the derivation and applications of the thin lens formula in optical systems.
  • Investigate the relationship between wavelength, slit separation, and fringe spacing in double slit experiments.
USEFUL FOR

Students studying wave optics, physics educators explaining interference phenomena, and researchers exploring optical properties of materials.

JohnGaltis
Messages
18
Reaction score
0

Homework Statement


Consider a double slit experiment with a light source of 540nm. Slit separation is 4mm. A thin glass plate is placed in front of one of the slit,, resulting in each bright fringe moving to the location of the adjacent bright fringe. Find the minimum thickness of the glass.

Refractive index of the glass is 1.56. Ignore reflections caused by the glass.

Homework Equations


2nt= +- mλ (bright fringe)
2nt= +- m+½λ (dark fringe)

The Attempt at a Solution


I think I should be using the thin lens formula since they emphasised the thin glass plate as "thin". Bright fringe moves the distance of a dark fringe to take the position of the adjacent bright fringe.

It seems too simple to just use

t= (m+½λ)/2n

Is there something I am missing?
 
Physics news on Phys.org
Hello John,

No lens. But I don't see what minimum thickness you propose. m, t, n ?

By the way, I'd like to see a drawing of the situation before and after placing the glass plate. I have difficulty understanding what exactly happens with the 1st maximum (does it move to where the 2nd was before?) Then how can the 2nd maximum move to where the 3rd was instead of to where the 4th was ?
 
BvU said:
Hello John,

No lens. But I don't see what minimum thickness you propose. m, t, n ?

By the way, I'd like to see a drawing of the situation before and after placing the glass plate. I have difficulty understanding what exactly happens with the 1st maximum (does it move to where the 2nd was before?) Then how can the 2nd maximum move to where the 3rd was instead of to where the 4th was ?
It's a conceptual question though.

mkbFoV4.jpg
 
I merely used:

Thickness= λ/4n

to obtain 8.65×10^-8 m
 
Bit thin for a glass plate, I would say...

What is the agrgument for your ##\lambda/(4n) ## ?

PS I initially misread a part. Perhaps this helps to compensate ...
 
Hmm but I don't have the distance between the slits and the wall.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
3
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 7 ·
Replies
7
Views
8K
  • · Replies 14 ·
Replies
14
Views
3K