हिंदी

In Young's double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. - Physics

Advertisements
Advertisements

प्रश्न

In Young's double slit experiment using light of wavelength 600 nm, the slit separation is 0.8 mm and the screen is kept 1.6 m from the plane of the slits. Calculate

  1. the fringe width
  2. the distance of (a) third minimum and (b) fifth maximum, from the central maximum.
संख्यात्मक

उत्तर

(i) Fringe width = β = `(λ"D")/"d"`

β = `(600 xx 10^-9 xx 1.6)/(0.8 xx 10^-3)`

∴ β = 12 × 10−4 m

(ii)

  1. Distance of 3rd minimum from the central maximum = `5/2`β = 30 × 10−4 m
  2. Distance of 5th maximum from central fringe = 5β = 60 × 10−4 m
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2021-2022 (April) Term 2 - Outside Delhi Set 1

संबंधित प्रश्न

Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500 nm. The separation between the slits is \[2 \cdot 0 \times  {10}^{- 3}m.\]


A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?


In a Young's double slit experiment, using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 micron (1 micron = 10−6 m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.


In a Young's double slit experiment, \[\lambda = 500\text{ nm, d = 1.0 mm and D = 1.0 m.}\] Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.


Draw a neat labelled diagram of Young’s Double Slit experiment. Show that `beta = (lambdaD)/d` , where the terms have their usual meanings (either for bright or dark fringe).


In Young’s double slit experiment, what is the effect on fringe pattern if the slits are brought closer to each other?


How will the interference pattern in Young's double-slit experiment be affected if the phase difference between the light waves emanating from the two slits S1 and S2 changes from 0 to π and remains constant?


Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light source (λ = 632.8 nm). The distance between the screen and the slits is 100 cm. If a bright fringe is observed on a screen at distance of 1.27 mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to :


The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of the fourth bright fringe after a glass plate of refractive index 1.5 is introduced in the path of one of the beams. The thickness of the glass plate would be ______.


The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×