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Two slits separated by a distance of 1mm are illuminated with red light of wavelength 6.5×10^-7m.the interference fringes are observed on a screen placed 1m from the slits. The distance between the third dark fringe n the fifth bright fringe on the same side of central maxima is 1) 0.65mm 2)1.62mm 3)3.25mm 4)4.88mm?
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Two slits separated by a distance of 1mm are illuminated with red ligh...
Given:
- Distance between the two slits (d) = 1mm = 0.001m
- Wavelength of red light (λ) = 6.5×10^-7m
- Distance of the screen from the slits (D) = 1m

To find:
The distance between the third dark fringe and the fifth bright fringe on the same side of the central maximum.

Solution:
1. Finding the distance between adjacent fringes:
The distance between adjacent bright and dark fringes can be given by the formula:
D = λD / d

Where, D is the distance between adjacent fringes.

Substituting the given values, we have:
D = (6.5×10^-7m x 1m) / (0.001m)
= 6.5×10^-4m

2. Finding the distance between the third dark fringe and the fifth bright fringe:
The distance between the third dark fringe and the fifth bright fringe can be calculated by using the formula:
Distance = (n + 1/2) x D

Where, n is the number of fringes.

For the third dark fringe, n = 3 and for the fifth bright fringe, n = 5.

Substituting the values in the formula, we get:
Distance = (3 + 1/2) x 6.5×10^-4m
= (7/2) x 6.5×10^-4m
= 3.25×10^-3m
= 3.25mm

Therefore, the distance between the third dark fringe and the fifth bright fringe on the same side of the central maximum is 3.25mm.
So, the correct option is 3) 3.25mm.
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Read the following text and answer the following questions on the basis of the same:Electron Microscope Electron microscopes use electrons to illuminate a sample. In Transmission Electron Microscopy (TEM), electrons pass through the sample and illuminate film or a digital camera.Resolution in microscopy is limited to about half of the wavelength of the illumination source used to image the sample. Using visible light the best resolution that can be achieved by microscopes is about ~200 nm. Louis de Broglie showed that every particle or matter propagates like a wave. The wavelength of propagating electrons at a given accelerating voltage can be determined byThus, the wavelength of electrons is calculated to be 3.88 pm when the microscope is operated at 100 keV, 2. 74 pm at 200 keV and 2.24 pm at 300 keV. However, because the velocities of electrons in an electron microscope reach about 70% the speed of light with an accelerating voltage of 200 keV, there are relativistic effects on these electrons. Due to this effect, the wavelength at 100 keV, 200 keV and 300 keV in electron microscopes is 3.70 pm, 2.51 pm and 1.96 pm, respectively.Anyhow, the wavelength of electrons is much smaller than that of photons (2.5 pm at 200 keV). Thus if electron wave is used to illuminate the sample, the resolution of an electron microscope theoretically becomes unlimited. Practically, the resolution is limited to ~0.1 nm due to the objective lens system in electron microscopes. Thus, electron microscopy can resolve subcellular structures that could not be visualized using standard fluorescences microscopy.Q. Why electron as wave is used in electron microscope to illuminate the sample?

Read the following text and answer the following questions on the basis of the same:Electron Microscope Electron microscopes use electrons to illuminate a sample. In Transmission Electron Microscopy (TEM), electrons pass through the sample and illuminate film or a digital camera.Resolution in microscopy is limited to about half of the wavelength of the illumination source used to image the sample. Using visible light the best resolution that can be achieved by microscopes is about ~200 nm. Louis de Broglie showed that every particle or matter propagates like a wave. The wavelength of propagating electrons at a given accelerating voltage can be determined byThus, the wavelength of electrons is calculated to be 3.88 pm when the microscope is operated at 100 keV, 2. 74 pm at 200 keV and 2.24 pm at 300 keV. However, because the velocities of electrons in an electron microscope reach about 70% the speed of light with an accelerating voltage of 200 keV, there are relativistic effects on these electrons. Due to this effect, the wavelength at 100 keV, 200 keV and 300 keV in electron microscopes is 3.70 pm, 2.51 pm and 1.96 pm, respectively.Anyhow, the wavelength of electrons is much smaller than that of photons (2.5 pm at 200 keV). Thus if electron wave is used to illuminate the sample, the resolution of an electron microscope theoretically becomes unlimited. Practically, the resolution is limited to ~0.1 nm due to the objective lens system in electron microscopes. Thus, electron microscopy can resolve subcellular structures that could not be visualized using standard fluorescences microscopy.Q. In electron microscope, electron is used

Two slits separated by a distance of 1mm are illuminated with red light of wavelength 6.5×10^-7m.the interference fringes are observed on a screen placed 1m from the slits. The distance between the third dark fringe n the fifth bright fringe on the same side of central maxima is 1) 0.65mm 2)1.62mm 3)3.25mm 4)4.88mm?
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Two slits separated by a distance of 1mm are illuminated with red light of wavelength 6.5×10^-7m.the interference fringes are observed on a screen placed 1m from the slits. The distance between the third dark fringe n the fifth bright fringe on the same side of central maxima is 1) 0.65mm 2)1.62mm 3)3.25mm 4)4.88mm? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about Two slits separated by a distance of 1mm are illuminated with red light of wavelength 6.5×10^-7m.the interference fringes are observed on a screen placed 1m from the slits. The distance between the third dark fringe n the fifth bright fringe on the same side of central maxima is 1) 0.65mm 2)1.62mm 3)3.25mm 4)4.88mm? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two slits separated by a distance of 1mm are illuminated with red light of wavelength 6.5×10^-7m.the interference fringes are observed on a screen placed 1m from the slits. The distance between the third dark fringe n the fifth bright fringe on the same side of central maxima is 1) 0.65mm 2)1.62mm 3)3.25mm 4)4.88mm?.
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