Oblique beam on a double slit

JEE Advanced 2019 Paper 2, Question 7

In a Young’s double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle \alpha as shown in figure. On the screen, the point O is equidistant from the slits and distance P O is 11.0 mm. Which of the following statement(s) is/are correct?

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  1. For \alpha=\frac{0.36}{\pi} degree, there will be destructive interference at point O.
  2. For \alpha=0, there will be constructive interference at point P.
  3. For \alpha=\frac{0.36}{\pi} degree, there will be destructive interference at point P.
  4. Fringe spacing depends on \alpha.


This setup is different from the usual double slit experiment in that there is an extra phase shift introduced by the angle of the incident beam, \alpha. This introduces an extra path difference of d \sin \alpha between the two rays. Of course, there is also the path difference yd/D incurred after light passes through the slits. Therefore the two light wavees arrive at the screen with a total path difference of

(1)   \begin{equation*}   \Delta = d \sin \alpha + \frac{y d}{D} \end{equation*}

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If we express \alpha in radians, we can approximate d \sin \alpha \approx d \alpha for small values of \alpha. Again, note that it is crucial we represent \alpha in radians when we do this. Furthermore, we have constructive interference when the path difference is an integer multiple of full wavelengths, and it is destructive when it is a half-integer multiple. That is,

(2)   \begin{equation*}   \Delta \approx d \alpha + \frac{y d}{D} =   \begin{cases}     m \lambda, &\text{  (constructive)} \\     (m+\frac{1}{2})\lambda, &\text{  (destructive)} .   \end{cases} \end{equation*}

We can now consider each option given in the problem. First we note that \alpha = \frac{0.36}{\pi} degrees is

(3)   \begin{equation*}   \alpha = \frac{0.36}{\pi} \times \frac{\pi}{180} = 2 \times 10^{-3} \text{ radians} . \end{equation*}

At point O on the screen y=0, and by (2)

(4)   \begin{equation*}   \Delta     = 0.3 \text{ mm} \times 2 \times 10^{-3} \text{ rad}     = 600 \text{ nm}, \end{equation*}

which is exactly one wavelength (m=1), ensuring constructive interference at O. So option (A) is incorrect.

For \alpha=\frac{0.36}{\pi} degrees, at point P (y=11 \text{ mm}),

(5)   \begin{equation*}   \Delta     = 0.3 \text{ mm} \times 2 \times 10^{-3} \text{ rad } +       \frac{11 \text{ mm} \times 0.3 \text{ mm}}{1 \text{ m}}     = 3900 \text{ nm } \equiv 6.5 \lambda \end{equation*}

which means we have destructive interference at P. If \alpha were zero, we’d have \Delta = 3300 \equiv 5.5 \lambda, which would also mean destructive interference. So option (B) is incorrect, but option (C) is correct.

Finally, we note that the additional path difference d \alpha shifts all the fringes uniformly; the spacing between the fringes is still

(6)   \begin{equation*}   y_{m+1}-y_m = \frac{d}{D} , \end{equation*}

which is independent of \alpha. Therefore, option (D) is incorrect.

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