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The set-up is shown in the image above. Find the intensity on the screen as a function of $y$, and $I_0$, where $I_0$ is the intensity of the central maxima.

I recognize that the plane wavefront incident on the first pair of slits results in two cylindrical wavefronts, which interfere on the second pair of slits. After this, I'm a bit lost. Are two more cylindrical wave-fronts set up, with whatever amplitude and phase the first interference causes? How would one go about finding the intensity?

Also, what is meant by "central maxima"? If two cylindrical wavefronts are set up in the second pair of slits, I don't see why there should necessarily be a maxima in the center. Does the term merely refer to a point where the phase difference is zero?

For example, if a plane wavefront wasn't incident normally on pair of slits, and that resulted in no maxima being present on the axis of symmetry, would we call the point with zero phase difference the central maxima, or would we say that the central maxima doesn't exist?

The answer for the problem is $$I_0\cos^2\left(\frac{2\pi }{\lambda}\frac{d_1d_2}{D_1}\right)\cos^2\left(\frac{2\pi }{\lambda }\frac{d_2 y}{D_2}\right)$$

  • $\begingroup$ Can you give a reference to where you get this question from, especially the answer? $\endgroup$
    – Kksen
    Jul 30, 2021 at 3:03
  • $\begingroup$ @KamKahSen It is from Pathfinder, an Indian physics olympiad book. $\endgroup$
    – Aspirant
    Jul 30, 2021 at 5:33

1 Answer 1


There are various assumptions and approximations made that are likely stated in the text before the problem was assigned. The slits are approximated as being sources of outgoing waves of the same wavelenght with magnitude independent of direction of either the incoming or outgoing waves. In that case, taking the dotted line as $y=0$, and $y$ as the position of the measurement on the right-hand screen, there are four possible paths that the light can travel to reach the screen. These are

  1. upper left slit to upper right slit to screen with path length $\ell_1+\ell_3$,

  2. upper left slit to lower right slit to screen with path length $\ell_2+\ell_4$,

  3. lower left slit to upper right slit to screen with path length $\ell_2+\ell_3$,

  4. lower left slit to lower right slit to screen with path length $\ell_1+\ell_4$,

where from the geometry \begin{eqnarray} \ell_1 &=& \sqrt{\frac{(d_2-d_1)^2}{4}+D_1^2} \\ \ell_2 &=& \sqrt{\frac{(d_2+d_1)^2}{4}+D_1^2} \\ \ell_3 &=& \sqrt{\frac{(d_2-2y)^2}{4}+D_1^2} \\ \ell_4 &=& \sqrt{\frac{(d_2+2y)^2}{4}+D_1^2} \,. \end{eqnarray} The amplitude at the screen is then proportional to \begin{equation} A \propto e^{i\frac{2\pi}{\lambda}(\ell_1+\ell_3)} + e^{i\frac{2\pi}{\lambda}(\ell_2+\ell_4)} + e^{i\frac{2\pi}{\lambda}(\ell_2+\ell_3)} + e^{i\frac{2\pi}{\lambda}(\ell_1+\ell_4)} \end{equation} and the intensity is the magnitude of the amplitude squared \begin{equation} I = |A|^2 \propto \cos^2\frac{2\pi}{\lambda}\frac{\ell_2-\ell_1}{2} \cos^2\frac{2\pi}{\lambda}\frac{\ell_3-\ell_4}{2} \,. \end{equation}

In order for the approximations of the slits above to be valid, the spacing of the slits needs to be small compared to the separations. So $D_1,D_2 \gg d_1,d_2$, and expanding the square root \begin{eqnarray} \ell_1 = D_1+\frac{(d_2-d_1)^2}{8D_1}+... \nonumber\\ \ell_2 = D_1+\frac{(d_2+d_1)^2}{8D_1}+... \nonumber\\ \ell_3 = D_2+\frac{(d_2-2y)^2}{8D_2}+... \nonumber\\ \ell_4 = D_2+\frac{(d_2+2y)^2}{8D_2}+... \end{eqnarray} so that \begin{eqnarray} \frac{\ell_2-\ell_1}{2} = \frac{d_2d_1}{4D_1} +... \nonumber\\ \frac{\ell_3-\ell_4}{2} = \frac{d_2y}{2D_2} +... \end{eqnarray} This gives \begin{equation} I \propto \cos^2\left (\frac{2\pi}{\lambda} \frac{d_1d_2}{4D_1}\right) \cos^2\left (\frac{2\pi}{\lambda} \frac{d_2y}{2D_2}\right) \end{equation} The maximum is at $y=0$, which is normally what is called the central maximum. Calling this $I_0$, the result is \begin{equation} I = I_0 \cos^2\left (\frac{2\pi}{\lambda} \frac{d_2y}{2D_2}\right) \,. \end{equation} I may have made factor of 2 errors, and I'm too lazy to check, but still it appears to me that the answer you give is in error, and should not include the first $\cos^2$ term

  • $\begingroup$ Maybe we get two $\cos^2$ terms if we assume that the central maxima is $y=0$ on the first pair of slits? Also, could you please check your working? Errors in this book are very rare. $\endgroup$
    – Aspirant
    Jul 30, 2021 at 5:35
  • $\begingroup$ Yes,we can get that term if it is the situation you mentioned. I just downloaded the book and checked it,the central maximun refers to that on the first screen. $\endgroup$
    – Kksen
    Jul 30, 2021 at 6:04
  • $\begingroup$ My brief calculation is the same with user2000143 basically. Not sure what went wrong. $\endgroup$
    – Kksen
    Jul 30, 2021 at 6:52
  • $\begingroup$ @KamKahSen The book is well known for its unintuitive and non-standard problems. Maybe we are missing something. $\endgroup$
    – Aspirant
    Jul 31, 2021 at 4:51
  • $\begingroup$ Yes,there is a high chance it is the case,thanks for reminding. $\endgroup$
    – Kksen
    Jul 31, 2021 at 5:00

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