मराठी

In Young’S Double Slit Experiment, Show That: - Physics (Theory)

Advertisements
Advertisements

प्रश्न

In Young’s double-slit experiment, show that: 

`beta = (lambda "D")/"d"` where the terms have their usual meaning.

टीपा लिहा
बेरीज

उत्तर

d = Distance between the slits

D = Distance between slit and screen

'P' is the position of mth order bright fringe.

From diagram, the path difference p is

S2P - S1P = S2P - AP

= S2

`"sin"theta = ("S"_2"A")/("S"_1"S"_2)` [From Δ S1AS2]


`"tan" theta = "OP"/"CO"` [From Δ POC]

θ → 0 ∴ sin θ ≅ tan θ


∴ `("S"_2"A")/("S"_1"S"_2) = "OP"/"OC"`


`therefore ("S"_2"A")/"d" = "x"/"D"`


`therefore "S"_2"A" = "x d"/"D"`

or p = `"x" "d"/"D"`

∴ For bright fringe,


`"x d"/"D" = "m" lambda`, where m is an integer

∴ xm = `("m"lambda"D")/"d"`

∴ xm+1 = (m + 1) `(lambda"D")/"d"`

∴ Fringe width β = [(m + 1)-m] `(lambda"D")/"d"`

β = `(lambda"D")/"d"`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2015-2016 (March) Set 1

संबंधित प्रश्‍न

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.


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.


Using analytical method for interference bands, obtain an expression for path difference between two light waves.


In Young’s experiment, the ratio of intensity at the maxima and minima in an interference
pattern is 36 : 9. What will be the ratio of the intensities of two interfering waves?


The fringes produced in diffraction pattern are of _______.

(A) equal width with same intensity

(B) unequal width with varying intensity

(C) equal intensity\

(D) equal width with varying intensity


The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I0. If one of the slits is closed, the intensity at this point will be ____________ .


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?


A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?


Why is the diffraction of sound waves more evident in daily experience than that of light wave?


In Young’s double slit experiment, how is interference pattern affected when the following changes are made:

  1. Slits are brought closer to each other.
  2. Screen is moved away from the slits.
  3. Red coloured light is replaced with blue coloured light.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×