**JEE Advanced 2019 Paper 2, Question 7**

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

- For degree, there will be destructive interference at point .
- For , there will be constructive interference at point .
- For degree, there will be destructive interference at point .
- Fringe spacing depends on .

**Solution**

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, . This introduces an extra path difference of between the two rays. Of course, there is also the path difference incurred after light passes through the slits. Therefore the two light wavees arrive at the screen with a total path difference of

(1)

If we express in radians, we can approximate for small values of . Again, note that it is crucial we represent 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)

We can now consider each option given in the problem. First we note that *degrees* is

(3)

At point on the screen , and by (2)

(4)

which is exactly one wavelength (), ensuring constructive interference at . So **option (A) is incorrect**.

For degrees, at point (),

(5)

which means we have destructive interference at . If were zero, we’d have , which would also mean destructive interference. So **option (B) is incorrect**, but **option (C) is correct**.

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

(6)

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