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In Young's double slit experiment, monochromatic light of wavelength 5000Å is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.
Correct answer is '125'. Can you explain this answer?
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In Youngs double slit experiment, monochromatic light of wavelength 50...
Let intensity of light on screen due to each slit is I0
So internity at centre of screen is 4��0
Intensity at distance y from centre
So internity at centre of screen is 4��0
Intensity at distance y from centre
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In Youngs double slit experiment, monochromatic light of wavelength 50...
Understanding Young's Double Slit Experiment
In Young's double slit experiment, we analyze how light behaves when passing through two closely spaced slits, resulting in an interference pattern on a screen.
Given Parameters:
- Wavelength (λ): 5000 Å (which is 5000 × 10^-10 m)
- Distance between slits (d): 1.0 mm (which is 1.0 × 10^-3 m)
- Distance to the screen (L): 1.0 m
Intensity in Interference Patterns:
The intensity I at a point on the screen can be expressed in terms of maximum intensity (I0) and the phase difference (Δφ) between the waves coming from the two slits. The formula for intensity is:
I = I0 * cos²(Δφ/2)
Finding Half Maximum Intensity:
To find the point where intensity becomes half of the maximum intensity:
I = I0 / 2
Substituting this into the intensity formula gives:
I0 * cos²(Δφ/2) = I0 / 2
This simplifies to:
cos²(Δφ/2) = 1/2
The angle corresponding to this condition is:
Δφ/2 = π/4 → Δφ = π/2
Relationship Between Phase Difference and Position:
The phase difference Δφ can be related to the path difference (Δx) using:
Δφ = (2π/λ) * Δx
For small angles, the path difference Δx can be approximated as:
Δx = d * sin(θ) ≈ d * y/L
Substituting for Δφ:
(2π/λ) * (d * y/L) = π/2
Solving for y gives:
y = (λ * L) / (2 * d)
Calculating the Distance y:
Substituting the values:
y = (5000 × 10^-10 m * 1 m) / (2 * 1.0 × 10^-3 m)
Calculating this yields:
y = 125 × 10^-6 m
Thus, the distance from the center of the screen where intensity becomes half of the maximum intensity for the first time is 125 × 10^-6 m.
In Young's double slit experiment, we analyze how light behaves when passing through two closely spaced slits, resulting in an interference pattern on a screen.
Given Parameters:
- Wavelength (λ): 5000 Å (which is 5000 × 10^-10 m)
- Distance between slits (d): 1.0 mm (which is 1.0 × 10^-3 m)
- Distance to the screen (L): 1.0 m
Intensity in Interference Patterns:
The intensity I at a point on the screen can be expressed in terms of maximum intensity (I0) and the phase difference (Δφ) between the waves coming from the two slits. The formula for intensity is:
I = I0 * cos²(Δφ/2)
Finding Half Maximum Intensity:
To find the point where intensity becomes half of the maximum intensity:
I = I0 / 2
Substituting this into the intensity formula gives:
I0 * cos²(Δφ/2) = I0 / 2
This simplifies to:
cos²(Δφ/2) = 1/2
The angle corresponding to this condition is:
Δφ/2 = π/4 → Δφ = π/2
Relationship Between Phase Difference and Position:
The phase difference Δφ can be related to the path difference (Δx) using:
Δφ = (2π/λ) * Δx
For small angles, the path difference Δx can be approximated as:
Δx = d * sin(θ) ≈ d * y/L
Substituting for Δφ:
(2π/λ) * (d * y/L) = π/2
Solving for y gives:
y = (λ * L) / (2 * d)
Calculating the Distance y:
Substituting the values:
y = (5000 × 10^-10 m * 1 m) / (2 * 1.0 × 10^-3 m)
Calculating this yields:
y = 125 × 10^-6 m
Thus, the distance from the center of the screen where intensity becomes half of the maximum intensity for the first time is 125 × 10^-6 m.
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In Youngs double slit experiment, monochromatic light of wavelength 5000 is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.Correct answer is '125'. Can you explain this answer?
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In Youngs double slit experiment, monochromatic light of wavelength 5000 is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.Correct answer is '125'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In Youngs double slit experiment, monochromatic light of wavelength 5000 is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.Correct answer is '125'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In Youngs double slit experiment, monochromatic light of wavelength 5000 is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.Correct answer is '125'. Can you explain this answer?.
In Youngs double slit experiment, monochromatic light of wavelength 5000 is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.Correct answer is '125'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about In Youngs double slit experiment, monochromatic light of wavelength 5000 is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.Correct answer is '125'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In Youngs double slit experiment, monochromatic light of wavelength 5000 is used. The slits are 1.0 mm apart and screen is placed at 1.0 m away from slits. The distance from the centre of the screen where intensity becomes half of the maximum intensity for the first time is __ × 10−6 m.Correct answer is '125'. Can you explain this answer?.
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