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In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :
- a)9.75 mm
- b)15.6 mm
- c)1.56 mm
- d)7.8 mm
Correct answer is option 'D'. Can you explain this answer?
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In a Young's double slit experiment, slits are separated by 0.5 mm...
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In a Young's double slit experiment, slits are separated by 0.5 mm...
Given data:
- Distance between slits: 0.5 mm
- Distance to the screen: 150 cm
- Wavelength 1: 650 nm
- Wavelength 2: 520 nm
To find: The least distance from the common central maximum to the point where the bright fringes due to both wavelengths coincide.
Derivation:
1. The condition for constructive interference in Young's double-slit experiment is given by the formula:
dsinθ = mλ
where d is the distance between the slits, θ is the angle between the central maximum and the fringe, m is the order of the fringe, and λ is the wavelength of the light.
2. In order to determine the distance between the central maximum and the point where the bright fringes coincide, we need to find the difference in path lengths for both wavelengths.
3. The path difference (Δx) between the two wavelengths can be expressed as:
Δx = dsinθ1 - dsinθ2
where θ1 and θ2 are the angles of the two wavelengths.
4. As we want the least distance, we consider the condition for constructive interference, i.e., m = 0.
5. For the central maximum, θ1 = θ2 = 0. Therefore, sinθ1 = sinθ2 = 0.
6. Substituting these values into the path difference formula, we get:
Δx = d(0) - d(0) = 0
7. Hence, the bright fringes due to both wavelengths coincide at the central maximum.
8. The distance from the central maximum to the bright fringe due to both wavelengths coinciding is given by the formula:
x = Rtanθ
where R is the distance to the screen and θ is the angle between the central maximum and the fringe.
9. To find θ, we can use the small angle approximation:
θ = tanθ ≈ sinθ = λ/R
where λ is the average of the two wavelengths.
10. Substituting the values, we have:
θ = (650 nm + 520 nm) / (2 * 150 cm) = 1170 nm / 300 cm = 39 nm / 10 cm = 3.9 nm / cm
11. Finally, substituting the value of θ into the formula for x, we get:
x = Rtanθ = (150 cm)(3.9 nm / cm) = 585 nm = 0.585 mm
Therefore, the least distance from the common central maximum to the point where the bright fringes due to both wavelengths coincide is 0.585 mm, which is approximately equal to 0.6 mm.
- Distance between slits: 0.5 mm
- Distance to the screen: 150 cm
- Wavelength 1: 650 nm
- Wavelength 2: 520 nm
To find: The least distance from the common central maximum to the point where the bright fringes due to both wavelengths coincide.
Derivation:
1. The condition for constructive interference in Young's double-slit experiment is given by the formula:
dsinθ = mλ
where d is the distance between the slits, θ is the angle between the central maximum and the fringe, m is the order of the fringe, and λ is the wavelength of the light.
2. In order to determine the distance between the central maximum and the point where the bright fringes coincide, we need to find the difference in path lengths for both wavelengths.
3. The path difference (Δx) between the two wavelengths can be expressed as:
Δx = dsinθ1 - dsinθ2
where θ1 and θ2 are the angles of the two wavelengths.
4. As we want the least distance, we consider the condition for constructive interference, i.e., m = 0.
5. For the central maximum, θ1 = θ2 = 0. Therefore, sinθ1 = sinθ2 = 0.
6. Substituting these values into the path difference formula, we get:
Δx = d(0) - d(0) = 0
7. Hence, the bright fringes due to both wavelengths coincide at the central maximum.
8. The distance from the central maximum to the bright fringe due to both wavelengths coinciding is given by the formula:
x = Rtanθ
where R is the distance to the screen and θ is the angle between the central maximum and the fringe.
9. To find θ, we can use the small angle approximation:
θ = tanθ ≈ sinθ = λ/R
where λ is the average of the two wavelengths.
10. Substituting the values, we have:
θ = (650 nm + 520 nm) / (2 * 150 cm) = 1170 nm / 300 cm = 39 nm / 10 cm = 3.9 nm / cm
11. Finally, substituting the value of θ into the formula for x, we get:
x = Rtanθ = (150 cm)(3.9 nm / cm) = 585 nm = 0.585 mm
Therefore, the least distance from the common central maximum to the point where the bright fringes due to both wavelengths coincide is 0.585 mm, which is approximately equal to 0.6 mm.
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In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :a)9.75 mmb)15.6 mmc)1.56 mmd)7.8 mmCorrect answer is option 'D'. Can you explain this answer?
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In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :a)9.75 mmb)15.6 mmc)1.56 mmd)7.8 mmCorrect answer is option 'D'. 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 a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :a)9.75 mmb)15.6 mmc)1.56 mmd)7.8 mmCorrect answer is option 'D'. 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 a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :a)9.75 mmb)15.6 mmc)1.56 mmd)7.8 mmCorrect answer is option 'D'. Can you explain this answer?.
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :a)9.75 mmb)15.6 mmc)1.56 mmd)7.8 mmCorrect answer is option 'D'. 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 a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :a)9.75 mmb)15.6 mmc)1.56 mmd)7.8 mmCorrect answer is option 'D'. 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 a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :a)9.75 mmb)15.6 mmc)1.56 mmd)7.8 mmCorrect answer is option 'D'. Can you explain this answer?.
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