A double slit setup is shown in the figure One of the slits is in medium 2 of refractive index $n_2$. The other slit is at the interface of this medium with another medium 1 of refractive index $n_1\left(\neq n_2\right)$. 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 $\theta$ from the line joining them, so that $\theta$ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.
Which of the following statement(s) is (are) correct?
Which of the following statement(s) is (are) correct?
- The phase difference between the two rays is independent of $d$.
- The two rays interfere constructively at the detector.
- The phase difference between the two rays depends on $n _1$ but is independent of $n_2$.
- The phase difference between the two rays vanishes only for certain values of $d$ and the angle of incidence of the beam, with $\theta$ being the corresponding angle of refraction.
The Correct Option is A, B
Solution and Explanation
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.
Top Questions on Youngs double slit experiment
- In Young's double slit experiment, the screen is moved 30 cm towards the slits. As a consequence, the fringe width of the pattern changes by 0.09 mm. If the slits separation used is 2 mm, calculate the wavelength of light used in the experiment.
- CBSE CLASS XII - 2025
- Physics
- Youngs double slit experiment
- Two wires A and B are made of the same material, having the ratio of lengths $ \frac{L_A}{L_B} = \frac{1}{3} $ and their diameters ratio $ \frac{d_A}{d_B} = 2 $. If both the wires are stretched using the same force, what would be the ratio of their respective elongations?
- JEE Main - 2025
- Physics
- Youngs double slit experiment
- A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is $ P \times 10^{11} \, \text{N/m}^2 $, where the value of $ P $ is: (Take $ g = 3\pi \, \text{m/s}^2 $)
- JEE Main - 2025
- Physics
- Youngs double slit experiment
- The two surfaces of a biconvex lens are of radius of curvature \( R \) each. Obtain the condition under which its focal length \( f \) is equal to \( R \). If one of the two surfaces of this lens is made plane, what will be the new focal length of the lens?
- CBSE CLASS XII - 2025
- Physics
- Youngs double slit experiment
- A light wave of wavelength 600 nm passes through a double-slit apparatus with a slit separation of 0.2 mm. What is the angular separation (in degrees) of the first-order bright fringe?
- JEECUP - 2025
- Physics
- Youngs double slit experiment
Questions Asked in JEE Advanced exam
As shown in the figures, a uniform rod $ OO' $ of length $ l $ is hinged at the point $ O $ and held in place vertically between two walls using two massless springs of the same spring constant. The springs are connected at the midpoint and at the top-end $ (O') $ of the rod, as shown in Fig. 1, and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $ f_1 $. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2, and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $ f_2 $. Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:
- JEE Advanced - 2025
- Waves and Oscillations
The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ____.
Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80- JEE Advanced - 2025
- Organic Chemistry
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
- JEE Advanced - 2025
- binomial expansion formula
- The total number of real solutions of the equation $$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1} \left( \frac{6 \tan \theta}{9 + \tan^2 \theta} \right) $$ is
(Here, the inverse trigonometric functions $ \sin^{-1} x $ and $ \tan^{-1} x $ assume values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $(-\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)- JEE Advanced - 2025
- Inverse Trigonometric Functions
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
- JEE Advanced - 2025
- Coordinate Geometry
Concepts Used:
Young’s Double Slit Experiment
- Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
- The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
- Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.
Read More: Young’s Double Slit Experiment