English

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.

Advertisements
Advertisements

Question

(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.

Advertisements

Solution

 

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

shaalaa.com
  Is there an error in this question or solution?
2015-2016 (March) Delhi Set 1

RELATED QUESTIONS

In Young' s experiment the ratio of intensity at the maxima and minima . in the interference pattern is 36 : 16. What is the ratio of the widths of the two slits?


In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.


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. 


In Young’s double slit experiment to produce interference pattern, obtain the conditions for constructive and destructive interference. Hence deduce the expression for the fringe width.


Two coherent sources of light having intensity ratio 81 : 1 produce interference fringes. Calculate the ratio of intensities at the maxima and minima in the interference pattern.


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 Young's double slit experiment shown in figure S1 and S2 are coherent sources and S is the screen having a hole at a point 1.0 mm away from the central line. White light (400 to 700 nm) is sent through the slits. Which wavelength passing through the hole has strong intensity?


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).


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.

In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×