To solve this problem, let's analyze the double slit setup and the conditions under which interference occurs. The setup involves two slits, one in medium 1 with refractive index \( n_1 \) and the other in medium 2 with refractive index \( n_2 \). The light incident on the slits is a monochromatic parallel beam of light, and it refracts when passing through the slits, with an angle of refraction \( \theta \) at the interface between the two media. The goal is to find which statement(s) are correct regarding the interference of the rays received by the detector.
1. Understanding the setup:
In the figure, light passes through the slits from medium 1 (with refractive index \( n_1 \)) into medium 2 (with refractive index \( n_2 \)). The angle of refraction \( \theta \) of the light is determined by the refractive indices of the two media. This setup is typical of an interference pattern, where the light passing through two slits will form an interference pattern on the detector placed in medium 2, depending on the phase difference between the two rays.
2. The Phase Difference Between the Two Rays:
In this setup, the phase difference between the two rays depends on several factors, including the path difference between them, which is influenced by the distance between the slits \( d \). However, the question specifically mentions that the angle \( \theta \) is the angle of refraction, which is determined by the refractive indices and the angle of incidence of the light. The phase difference at the detector is thus influenced by the refractive index of medium 1 (not medium 2), since the angle of refraction depends on \( n_1 \) (and not \( n_2 \)). Therefore, the phase difference between the two rays is independent of \( d \), the distance between the slits.
3. Interference at the Detector:
When the two rays recombine at the detector, interference occurs. Since both rays are coming from the slits in medium 1 and refracted at the interface between medium 1 and medium 2, they will interfere constructively or destructively depending on their phase difference. However, in this case, since the problem suggests that \( \theta \) is the angle of refraction, we know that the path difference and phase difference will be appropriate for constructive interference at the detector.
4. Conclusion:
From the analysis, we conclude the following:
The phase difference between the two rays is independent of the slit separation \( d \) because the refractive index of medium 1 determines the path difference (not \( d \)).
The two rays interfere constructively at the detector, as the phase difference allows for constructive interference.
The phase difference depends on \( n_1 \), the refractive index of medium 1, but is independent of \( n_2 \).
The phase difference between the two rays vanishes only for certain values of \( d \) and \( \theta \), corresponding to constructive interference.
Thus, the correct answers are A, B, and C.
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
Read More: Young’s Double Slit Experiment