मराठी

In Young'S Double-slit Experiment, Deduce the Condition for (A) Constructive and (B) Destructive Interferences at a Point on the Screen. Draw a Graph Showing Variation of Intensity in the Interference Pattern Against Position 'X' on the Screen.

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प्रश्न

(i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interferences at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.

(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.

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उत्तर

 

Expression for Fringe Width in Young’s Double-Slit Experiment

Let S1 and S2 be two slits separated by a distance d. `GG'` is the screen at a distance D from the slits S1and S2. Point C is equidistant from both the slits. The intensity of light will be maximum at this point because the path difference of the waves reaching this point will be zero.

At point P, the path difference between the rays coming from the slits is given by

S1= S2P  − S1P

Now, S1 S2 = d, EF = d, and S2F = D

In ΔS2PF,

`S_2P=[S_2F^2+PF^2]^(1/2)`

 
`S_2P=[D2+(x+d/2)^2]^(1/2)`

 

`=D[1+(x+d/2)^2/D^2]^(1/2)`

Similarly, in ΔS1PE

`S_1P=D[1+(x-d/2)^2/D^2]^(1/2)`

 

`:.S_2P-S_1P=D[1+1/2(x+d/2)^2/D^2)]-D[1+1/2(x-d/2)^2/D^2]`

On expanding it binomially, we get

`S_2P-S_1P=1/(2D)[4xd/2]=(xd)/D`

 For constructive interference, the path difference is an integral multiple of wavelengths, that is, path difference is .

`:.nlambda=(xd)/D`

`x=(nlambdaD)/d`

where n = 0, 1, 2, 3, 4, …

Similarly, for destructive interference,

`x_n=(2n-1)lambda/2D/d`

Graph of Intensity Distribution in Young’s Double-Slit Experiment

 

(ii) On comparing the interference pattern observed in Young's double slit experiment (interference) with single-slit diffraction pattern (diffraction), we can have three distinguishing features:

In the interference pattern, all the bright fringes have the same intensity. In a diffraction pattern, all the bright fringes are not of the same intensity

In the interference pattern, the dark fringe has zero or very small intensity so that the bright and dark fringes can easily be distinguished. In diffraction pattern, all the dark fringes are not of zero intensity

In the interference pattern, the widths of all the fringes are almost the same, whereas in diffraction pattern, the fringes are of different widths

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2015-2016 (March) Delhi Set 1

संबंधित प्रश्‍न

What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same?


 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.


How does the fringe width get affected, if the entire experimental apparatus of Young is immersed in water?


In a double slit interference experiment, the separation between the slits is 1.0 mm, the wavelength of light used is 5.0 × 10−7 m and the distance of the screen from the slits is 1.0m. (a) Find the distance of the centre of the first minimum from the centre of the central maximum. (b) How many bright fringes are formed in one centimetre width on the screen?


A source emitting light of wavelengths 480 nm and 600 nm is used in a double-slit interference experiment. The separation between the slits is 0.25 mm and the interference is observed on a screen placed at 150 cm from the slits. Find the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths.


What should be the path difference between two waves reaching a point for obtaining constructive interference in Young’s Double Slit experiment ?


"If the slits in Young's double slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit".

Justify the above statement through a relevant mathematical expression.


The force required to double the length of a steel wire of area 1 cm2, if its Young's modulus Y= 2 × 1011/m2 is: 


A slit of width 0.6 mm is illuminated by a beam of light consisting of two wavelengths 600 nm and 480 nm. The diffraction pattern is observed on a screen 1.0 m from the slit. Find:

  1. The distance of the second bright fringe from the central maximum pertaining to the light of 600 nm.
  2. The least distance from the central maximum at which bright fringes due to both wavelengths coincide.

In Young's double slit experiment, show that:

`β = (λ"D")/"d"`

Where the terms have their usual meaning.


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