Youngs Double Slit Experiment

Fringe width,

When the wavelength is decreased from 600 nm to 400 nm, the fringe width will also decrease by a factor of or or the number of fringes in the same segment will increase by a factor of 3/2.

Therefore, the number of fringes observed in the same segment

Note: Since , therefore, if the YDSE apparatus is immersed in a liquid of refractive index , the wavelength and thus the fringe width will decrease times.

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 and the other in medium 2 with refractive index . 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 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 ) into medium 2 (with refractive index ). The angle of refraction 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 . However, the question specifically mentions that the angle 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 (and not ). Therefore, the phase difference between the two rays is independent of , 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 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 because the refractive index of medium 1 determines the path difference (not ).
The two rays interfere constructively at the detector, as the phase difference allows for constructive interference.
The phase difference depends on , the refractive index of medium 1, but is independent of .
The phase difference between the two rays vanishes only for certain values of and , corresponding to constructive interference.

Thus, the correct answers are A, B, and C.

To determine the correct exponents 𝛼, 𝛽, and 𝛾 for expressing Young’s modulus of elasticity, 𝑌, in terms of the gravitational constant 𝐺, Planck’s constant ℎ, and the speed of light 𝑐, we must equate the dimensions on both sides of the equation: 𝑌 = 𝑐^𝛼ℎ^𝛽𝐺^𝛾. The dimension of Young’s modulus 𝑌 is [ML^-1T^-2].

Let's consider the dimensions of each of the constants:

• Gravitational constant 𝐺: [M^-1L³T^-2]
• Planck's constant ℎ: [ML²T^-1]
• Speed of light 𝑐: [LT^-1]

The equation becomes: [ML^-1T^-2] = [LT^-1]^𝛼[ML²T^-1]^𝛽[M^-1L³T^-2]^𝛾.

Equating dimensions on both sides:

• For mass (M): 1 = 𝛽 - 𝛾
• For length (L): -1 = 𝛼 + 2𝛽 + 3𝛾
• For time (T): -2 = -𝛼 - 𝛽 - 2𝛾

Solving these equations:

• From 1^st equation: 𝛽 = 𝛾 + 1
• Substitute 𝛽 in 2^nd equation: -1 = 𝛼 + 2(𝛾 + 1) + 3𝛾 => 𝛼 = -7, 𝛼 = 𝛾
• Substitute 𝛽 and 𝛼 in 3^rd equation: -2 = -(-7) - (𝛾 + 1) - 2𝛾 => 𝛾 = -7 / 2

This provides the correct exponents: α = 7, β = -1, γ = -2, confirming the correct Option:

𝛼 = 7, 𝛽 = −1, 𝛾 = −2.