Interference pattern due to a thin prism and a lens

Question

We have a setup like this where a converging lens placed between inclined face of thin glass wedge and a screen. The screen is parallel to the inclined face of the wedge and the optical axis of the lens is perpendicular to the screen as well as to the inclined face of the wedge. Parallel beam of light is incident on wedge. Distance of the lens from the inclined face is $a$ = 10 $mathrm{cm}$, that from the screen is $b$ = 100 $mathrm{cm}$, refractive index of the glass of the wedge is $mu$ = 1.5 and wavelength of light used is $lambda$ = 6000 Å .
If we have been given the width of the interference fringes on the screen as 1 $mathrm{mm}$, then how do we calculate the wedge angle $theta$ ?
My doubt:
All the paralell rays incident on the glass wedge will be deviated by an angle $delta=(mu-1)theta$. Since all these deviated rays also are paralell, the converging lens will focus them in the focal plane which is the screen here. With what two rays does the interference happen then? How to calculate the points of maxima and minima and the hence the fringe width here?

flippiefanus · Answer

If I understand the setup correctly, then the "fringes" you'll see would not be due to interference but because of different reflected orders passing through the lens at different angle producing different spots on the screen. The first beam enters the lens at an angle $theta_1$ given according to Snell's law by
$$ sin(theta_1)=nsin(theta) . $$
The second beam enters the lens at an angle $theta_2$ after having reflected back and forth once inside the prism to produce the angle by
$$ sin(theta_2)=nsin(3theta) . $$
The two spots on the screen are then separated by a distance
$$ Delta x = b tan(theta_2)- btan(theta_1) . $$
If these angles are small, we can assume $tan(theta_n)approxsin(theta_n)approxtheta_n$. Then we get
$$ Delta x approx b sin(theta_2)- bsin(theta_1) approx 2 b ntheta . $$

Interference pattern due to a thin prism and a lens

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