Double Slit Question with Thin Lens concept

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Homework Help Overview

The discussion revolves around a double slit experiment involving a light source of 540nm and a slit separation of 4mm. A thin glass plate is introduced in front of one of the slits, causing the bright fringes to shift to the position of adjacent bright fringes. Participants are tasked with determining the minimum thickness of the glass plate, given its refractive index of 1.56, while ignoring reflections.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the relevance of the thin lens formula and the implications of the glass plate being "thin." There are questions about the relationship between the fringe movement and the minimum thickness of the glass, with some suggesting that the optical path difference corresponds to a wavelength shift. Others express confusion about the movement of the maxima and seek clarification on the situation's setup.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have proposed calculations for the glass thickness, while others are questioning the assumptions and seeking visual aids to better understand the scenario. There is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants note the absence of information regarding the distance between the slits and the wall, which may affect the analysis. Additionally, there is mention of a related thread that could provide further context to the discussion.

JohnGaltis
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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?
 
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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.
 

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