मराठी

In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is 1/8 th of the wavelength.

Advertisements
Advertisements

प्रश्न

In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is `1/8`th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to ______.

पर्याय

  • 0.80

  • 0.74

  • 0.94

  • 0.85

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

In a Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is `1/8`th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to 0.85.

Explanation:

Δx = `λ/8`

Path difference `Δphi = (2π)/λ xx Δx  = (2π)/λ xx λ/8 = π/4`

`I = I_0 cos^2 ((Δphi)/2) = I_0 cos^2(π/8)`

`I/I_0 = cos^2(π/8)` = 0.85

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2022-2023 (March) Sample

संबंधित प्रश्‍न

In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is `λ/3`.


In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.


Find the intensity at a point on a screen in Young's double slit experiment where the interfering waves have a path difference of (i) λ/6, and (ii) λ/2. 


A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1 · 4 m away. If the two slits are separated by 0·28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.


 What is the effect on the interference fringes to a Young’s double slit experiment when

(i) the separation between the two slits is decreased?

(ii) the width of a source slit is increased?

(iii) the monochromatic source is replaced by a source of white light?

Justify your answer in each case.


A Young's double slit experiment is performed with white light.

(a) The central fringe will be white.

(b) There will not be a completely dark fringe.

(c) The fringe next to the central will be red.

(d) The fringe next to the central will be violet.


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).


Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit.


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 :


  • Assertion (A): In Young's double slit experiment all fringes are of equal width.
  • Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×