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A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
- a)1.37 mm
- b)1.27 mm
- c)1.47 mm
- d)9.75 mm
Correct answer is option 'D'. Can you explain this answer?
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Understanding Young's Double-Slit Experiment
In a Young’s double-slit experiment, the interference pattern is created by the superposition of light waves emerging from two closely spaced slits. The position of the bright fringes on the screen can be calculated using the formula:
Position of Bright Fringe
- The position of the m-th bright fringe is given by:
y = (m * λ * L) / d
Where:
- y = distance of the m-th bright fringe from the central maximum
- m = fringe order (0, 1, 2, 3,...)
- λ = wavelength of light
- L = distance from the slits to the screen
- d = distance between the slits
Given Data
- Wavelength (λ) for 650 nm = 650 x 10^-9 m
- Distance from slits to screen (L) = 1.4 m
- Distance between slits (d) = 0.28 mm = 0.28 x 10^-3 m
- We want to find the distance for the third bright fringe (m = 3).
Calculating the Distance
1. Substitute the values into the formula:
y = (3 * (650 x 10^-9) * 1.4) / (0.28 x 10^-3)
2. Calculate step-by-step:
- 3 * 650 x 10^-9 = 1950 x 10^-9 m
- 1950 x 10^-9 * 1.4 = 2730 x 10^-9 m
- Finally, divide by 0.28 x 10^-3 m:
y = (2730 x 10^-9) / (0.28 x 10^-3) = 9.75 x 10^-3 m = 9.75 mm
Conclusion
Thus, the distance of the third bright fringe from the central maximum for the wavelength of 650 nm is 9.75 mm, which corresponds to option 'D'.
In a Young’s double-slit experiment, the interference pattern is created by the superposition of light waves emerging from two closely spaced slits. The position of the bright fringes on the screen can be calculated using the formula:
Position of Bright Fringe
- The position of the m-th bright fringe is given by:
y = (m * λ * L) / d
Where:
- y = distance of the m-th bright fringe from the central maximum
- m = fringe order (0, 1, 2, 3,...)
- λ = wavelength of light
- L = distance from the slits to the screen
- d = distance between the slits
Given Data
- Wavelength (λ) for 650 nm = 650 x 10^-9 m
- Distance from slits to screen (L) = 1.4 m
- Distance between slits (d) = 0.28 mm = 0.28 x 10^-3 m
- We want to find the distance for the third bright fringe (m = 3).
Calculating the Distance
1. Substitute the values into the formula:
y = (3 * (650 x 10^-9) * 1.4) / (0.28 x 10^-3)
2. Calculate step-by-step:
- 3 * 650 x 10^-9 = 1950 x 10^-9 m
- 1950 x 10^-9 * 1.4 = 2730 x 10^-9 m
- Finally, divide by 0.28 x 10^-3 m:
y = (2730 x 10^-9) / (0.28 x 10^-3) = 9.75 x 10^-3 m = 9.75 mm
Conclusion
Thus, the distance of the third bright fringe from the central maximum for the wavelength of 650 nm is 9.75 mm, which corresponds to option 'D'.
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Question Description
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer?.
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer?.
Solutions for A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. The distance between the slit and the screen is 1.4m and the distance between the slits is 0.28mm. Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.a)1.37 mmb)1.27 mmc)1.47 mmd)9.75 mmCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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