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Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this?
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Plastic sheet refractive index is equals to 1.6 cover 1 slit of a doub...
Calculation of Minimum Thickness of Plastic Sheet
The phenomenon described in the question is known as the dark fringe in Young's double-slit experiment. In this case, the plastic sheet with a refractive index of 1.6 is placed in front of one of the slits.
Key Points:
- When the light passes through the plastic sheet, it undergoes a phase change of π due to the higher refractive index.
- This phase change leads to destructive interference at the center of the screen, resulting in a dark fringe instead of a bright one.
Formula Used:
The condition for destructive interference at the center of the screen is given by the formula:
\[t = \frac{\lambda}{2n}\]
where t is the thickness of the plastic sheet, λ is the wavelength of light (5867 Å = 586.7 nm), and n is the refractive index of the plastic sheet (1.6).
Calculation:
Substitute the values into the formula:
\[t = \frac{586.7}{2 \times 1.6} = 183.34 \text{ nm}\]
Therefore, the minimum thickness of the plastic sheet that needs to be used for the dark fringe to appear at the center of the screen is 183.34 nm.
The phenomenon described in the question is known as the dark fringe in Young's double-slit experiment. In this case, the plastic sheet with a refractive index of 1.6 is placed in front of one of the slits.
Key Points:
- When the light passes through the plastic sheet, it undergoes a phase change of π due to the higher refractive index.
- This phase change leads to destructive interference at the center of the screen, resulting in a dark fringe instead of a bright one.
Formula Used:
The condition for destructive interference at the center of the screen is given by the formula:
\[t = \frac{\lambda}{2n}\]
where t is the thickness of the plastic sheet, λ is the wavelength of light (5867 Å = 586.7 nm), and n is the refractive index of the plastic sheet (1.6).
Calculation:
Substitute the values into the formula:
\[t = \frac{586.7}{2 \times 1.6} = 183.34 \text{ nm}\]
Therefore, the minimum thickness of the plastic sheet that needs to be used for the dark fringe to appear at the center of the screen is 183.34 nm.
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Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this?
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Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this?.
Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Plastic sheet refractive index is equals to 1.6 cover 1 slit of a double slit arrangement for the young experiments when the doubleslit is eliminated by monochromatic like 5867 Armstrong the centre of the screen appear dark rather than bright the minimum thickness of the plastic sheet to be used for this?.
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