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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. - Physics

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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.

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

In Young's double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.


The ratio of the intensities at minima to the maxima in the Young's double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.


A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.


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


Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit.


Write two characteristics features distinguish the diffractions pattern from the interference fringes obtained in Young’s double slit experiment.


In a Young's double slit experiment, two narrow vertical slits placed 0.800 mm apart are illuminated by the same source of yellow light of wavelength 589 nm. How far are the adjacent bright bands in the interference pattern observed on a screen 2.00 m away?


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.


Young's double slit experiment is made in a liquid. The 10th bright fringe lies in liquid where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately


ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.

REASON (R): Fringe width is proportional to (d/D).


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