in the diffraction pattern due to a single slit, the width of the central maximum will be
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get in the diffraction pattern due to a single slit, the width of the central maximum will be from screen.
The angular width of the central maximum in a single slit diffraction pattern is 60^o . The width of the slit is 1 mu m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the centres of each slit.)
Click here👆to get an answer to your question ✍️ The angular width of the central maximum in a single slit diffraction pattern is 60^o . The width of the slit is 1 mu m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the centres of each slit.)
Question o
. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50cm from the slits. If the observed fringe width is 1cm, what is slit separation distance? (i.e., distance between the centres of each slit.)
A75μm
B100μm
C25μm
D50μm
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Updated on : 2022-09-05
Solution Verified by Toppr
Correct option is C)
Angular width of the central maxima θc =60 o Half angle θ= 2 60 o =30 o Given : b=10 −6 m Using sinθ= b λ Or sin30 o = 10 −6 λ ⟹ λ=0.5×10 −6 m
Given : D=50 cm=0.5 m
Fringe width β=1 cm=0.01 m
Using β= d λ D ∴ 0.01= 0.01 0.5×10 −6 ×0.5 =25μm Video Explanation
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[Solved] In the diffraction pattern obtained in a single slit experim
CONCEPT: Single Slit Experiment: The diffraction or bending phenomenon of light in which light from a coherent source interfere with itself and produce a dis
Home Physics Optics Diffraction Diffraction by a Single Slit
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In the diffraction pattern obtained in a single slit experiment the width of the central bright maximum is _____________ as the width of the other maxima.
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four times half one fourth twice
Answer (Detailed Solution Below)
Option 4 : twice
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CONCEPT:Single Slit Experiment: The diffraction or bending phenomenon of light in which light from a coherent source interfere with itself and produce a distinctive pattern on the screen called the diffraction pattern.In single slit diffraction, the light that passes through a single slit is on the order of the wavelength of the light.
The diffraction pattern on the screen will be at a distance D away from the slit.
A also shown in the figure of single-slit diffraction, intensity showing the central maximum to be wider and much more intense than those to the sides.
The central maximum is six times higher than other maxima on either side.
Width of maxims are found using the given formula:Width of other maxima than central maximum = λ/D
Width of central maximum = 2λ/D
where λ is the wavelength of light and D is the distance of the screen from the slit.
CALCULATION:In the single-slit experiment
Width of other maxima than central maximum = λ/D
Width of central maximum = 2λ/D
So the width of the central bright maximum is twice as the width of the other maxima.So the correct answer is option 4.
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More Diffraction Questions
Q1. Angular width of the central maxima in the Fraunhofer diffraction for λ = 6000 Å is θ0. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is,Q2. A light wave is traveling linearly in a medium of dielectric constant 4, incidents on the horizontal interface are separating the medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be : (Given : relative permeability of medium μr = 1)Q3. In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:Q4. The wavelength of an X-ray beam is 10 Å. The mass of a fictitious particle having the same energy as that of the X-ray photons isx3
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Q5. In Young’s double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :Q6. Pick the wrong answer in the context with rainbow.Q7. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (μwater = 4/3)Q8. In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed toIn the case of diffraction pattern due to a single slit angular width of central max can be increased by
In the case of diffraction pattern due to a single slit angular width of central max can be increased by
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In the case of diffraction pat...
In the case of diffraction pattern due to a single slit angular width of central max can be increased by
Updated On: 27-06-2022
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Text Solution Open Answer in App A
decreasing the wave length
B
increasing the slit width
C
increasing distance between slit and screen
D
decreasing the slit width.
Answer
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Guys, does anyone know the answer?