The image of a setup

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?

  • $\begingroup$ The only way I see for fringes to appear is due to the internal reflection. Try to interfere those that are immediately transmitted with those that suffer 2 reflections. That is also the principle of the use of wedged etalons. However I'm finding it hard to picture in my mind how I would draw the lines in this specific problem, as the lens changes things a bit. $\endgroup$ May 23, 2021 at 17:03
  • $\begingroup$ Can you please tell what is the purpose of lens here and what it does with the fringe width $\endgroup$ May 23, 2021 at 17:07
  • $\begingroup$ Not sure. I would need to get a bit deeper into this. But I would approach this problem by making a ray diagram of the whole problem, including the direct transmission and the first order secondary reflection $\endgroup$ May 23, 2021 at 17:15
  • 2
    $\begingroup$ the purpose of the lens is to focus the interfering parallel rays onto the screen; look up "Fraunhofer diffraction" $\endgroup$
    – hyportnex
    May 26, 2021 at 21:28
  • $\begingroup$ How do paralell rays interfere? $\endgroup$ May 27, 2021 at 14:38

1 Answer 1


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 $$ \sin(\theta_1)=n\sin(\theta) . $$ 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 $$ \sin(\theta_2)=n\sin(3\theta) . $$ The two spots on the screen are then separated by a distance $$ \Delta x = b \tan(\theta_2)- b\tan(\theta_1) . $$ If these angles are small, we can assume $\tan(\theta_n)\approx\sin(\theta_n)\approx\theta_n$. Then we get $$ \Delta x \approx b \sin(\theta_2)- b\sin(\theta_1) \approx 2 b n\theta . $$


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