A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index n₂. The other slit is at the interface of this medium with another medium 1 of refractive index n₁ (n₂). The line joining the slits is perpendicular to the interface and the distance between the slits is d. The slit widths are much smaller than d. A monochromatic parallel beam of light is incident on the slits from medium 1. A detector is placed in medium 2 at a large distance from the slits, and at an angle θ from the line joining them, so that θ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.

Which of the following statements (s) is (are) correct ?

Correct Options:
1. The phase difference between the two rays in independent of d.
2. The two rays interfere constructively at the detector.

Step-by-Step Explanation:

1. Understanding the Setup:
Let medium 1 have a refractive index _n1 and medium 2 have a refractive index _n2. The line joining the two slits (Slit 1 and Slit 2) is perpendicular to the plane interface between the two media. Slit 1 lies on the interface, and Slit 2 is located inside medium 2 at a distance d from Slit 1 along the perpendicular line.

2. Optical Path Difference Analysis:
Let light rays be incident from medium 1 at an angle of incidence $i$ relative to the normal (perpendicular to the interface, which is along the line joining the slits).
According to Snell's Law at the interface between medium 1 and medium 2:

$n_1 \sin i=n_2 \sin \theta$

where $\theta$ is the angle of refraction in medium 2.

The parallel incident rays travel through medium 1 before reaching Slit 1 and Slit 2. The extra path traveled by the ray going to Slit 2 in medium 1 before hitting the interface is related to the slit separation d.
Specifically, the path difference introduced in medium 1 before reaching the line perpendicular to the wavefront is given by:

$\Delta x_1=d \cos i$

So the corresponding optical path in medium 1 is:

After passing the slits, the rays travel in medium 2 at angle $\theta$ with respect to the line of slits. The path difference in medium 2 between the two parallel rays going to the far detector is:

$\Delta x_2=d \cos \theta$

So the corresponding optical path in medium 2 is:

3. Total Optical Path Difference:
The net optical path difference between ray 1 and ray 2 arriving at the detector is the difference between these two contributions:

$\Delta \text{OPD}=n_1 d \cos i-n_2 d \cos \theta$

Since the incident wave in medium 1 refracts into medium 2 at angle $\theta$, the components of the wavevectors parallel to the boundary interface are equal according to Snell's law:

For a wave refracted at an angle equal to the angle of observation $\theta$, the total phase difference between the two rays at the detector evaluates to zero ($\Delta \varphi =0$).
Since $\Delta \text{OPD}=0$, the phase difference $\Delta \varphi =\frac{2\pi }{\lambda }\Delta \text{OPD}=0$.

4. Conclusion:
- A phase difference of 0 means it is independent of d.
- A phase difference of 0 corresponds to maximum constructive interference at the detector.
Therefore, the correct options are that the phase difference between the two rays is independent of d and the two rays interfere constructively at the detector.