How Does a Thin Glass Plate Affect Young's Double Slit Experiment?

[brad sue]

Homework Statement

The Young's double slit experiment is modified, as shown below, by placing a thin parallel glass plate of thickness d and refractive index n over one of the slits. Here a is the slit width in the x direction and b is the slit separation also in the x direction. The system is excited with on-axis collimated laser light of wavelength. A lens of focal length F is placed immediately behind the slits and a screen is placed a distance F away from the lens, so that a set of well-defined interference fringes are visible on the screen. Ignore all phase changes from reflection and transmission in this problem.

(a) For the case where the glass plate is absent, derive an expression for the position, X2m, of the mth maximum on the screen.
(b) Derive an expression for the spatial frequency shift [(sin (θ’))/λ - (sin θ)/λ] that occurs when the glass slide is inserted in the position shown in the diagram. What important conclusion do you draw from your expression?
(c) Derive an expression for the lateral spatial shift X’2m -X2m on the screen as a result of inserting the glass slide.
(d) In which direction do the fringes shift?

The Attempt at a Solution

I solved this problem but I did not use the fresnel nor fraunhofer equations.
Do you think i should use those equations or concept?
It is the lens that confuses me. I know that when we have a lens, the output field is the Fourier transform of the input field.
I attached the picture.

Please, can I have your suggestions?
thank you

 

[Gordianus]

You don't need Faraunhofer or Fresnel equations with this problem.
The lens serves to bring the so-called "far-field" (Fraunhofer diffraction) to comfortable distance within the laboratory. Forget about it.
The glass slab is the key. Since the speed of light changes inside the glass, so does the wavelength. Thus, the phase shift, after traveling a distance d changes according the refraction index. This phase shift makes the fringes move up or down. Repeat the basic calculation, but take into account the refractive index of glass.

 

[brad sue]

Thank you
As you said the lens just allow the fringes to be seen at a focal length . For example if the distance to the screen was 2* focal length the fringes would appear blurry.