Home
Class 12
PHYSICS
Obtain the expression for fringe width i...

Obtain the expression for fringe width in the case of interference of light waves.

Text Solution

Verified by Experts

Consider Young.s double slit arrangement for obtaining interference fringes as shown in figure. `S_(1) and S_(2)` -two coherent sources (Yong.s double slits)
d= distance between slits
D- distance of screen from coherent sources/slits
O- center point on the screen and is equidistant from `S_(1) and S_(2)`

The path difference between the two light waves from `S_(1) and S_(2)` reaching the point O is zero. Thus the point O has maximum intensity. Consider a point P at a distance x from O.
The path difference between the light waves from `S_(1) and S_(2)` reaching the point P is, `delta = S_(2) P- S_(1)P`
From the figure, `(S_(2)P)^(2)= (S_(2)F)^(2)= (FP)^(2)= D^(2) + (x + (d)/(2))^(2)`
Similarly `(S_(1)P)^(2)= (S_(1)E)^(2)= (EP)^(2)= D^(2) + (x- (d)/(2))^(2)`
`therefore (S_(2)P)^(2)- (S_(1)P)^(2)= [D^(2) + (x+ (d)/(2))^(2)]- [D^(2) + (x + (d)/(2))^(2)] = [D^(2) + x^(2) + (d^(2))/(4) + 2(x) ((d)/(2))]- [D^(2) + x^(2) + (d^(2))/(4) + 2(x) ((d)/(2))]= 2xd`
`(S_(2)P-S_(1)P) (S_(2)P) + S_(1)P)= 2xd`
`(S_(2)P-S_(1)P)= (2xd)/((S_(2)P+ S_(1)P))`
Since P is very close to O and `d lt lt D`, therefore `(S_(2)P + S_(1)P) ~~ 2D`
Path difference, `(S_(2)P-S_(1)P) = (2xd)/(2D)= (xd)/(D)` ....(1)
Equation (1) represents the path difference between light waves from `S_(1) and S_(2)` superposing at the point P.
For constructive interference, `S_(2)P- S_(1)P= n lamda`, where, n= 0, 1, 2........
`(xd)/(D)= n lamda or x= n ((lamda D)/(d))`
i.e., The distance of then `n^(th)` bright fring from the centre O of the screen is `x_(0)=n ((lamda D)/(d))`
The distance of `(n +1)^(th)` bright fringe from the centre of the scrren is `x_(n+1)= (n+1) ((lamda D)/(d))`
The distance between the centers of any two consecutive bright fringes is called the fringe width of bright fringes. The fringe width is given by,
`beta= x_(n+1) -x_(n)= (n+1) ((lamda D)/(d))-n ((lamda D)/(d))= (lamda D)/(d)`
`therefore beta= (lamda D)/(d)`
Similarly for dark fringes also we obtain the same expression for fringe width.
Promotional Banner

Topper's Solved these Questions

  • ANNUAL EXAM QUESTION PAPER MARCH-2020

    SUNSTAR PUBLICATION|Exercise Part-C|8 Videos
  • ANNUAL EXAM QUESTION PAPER MARCH -2015

    SUNSTAR PUBLICATION|Exercise PART-A|37 Videos
  • II PUC PHYSICS (P.U. BOARD LATEST MODEL QUESTIONS PAPER -2)

    SUNSTAR PUBLICATION|Exercise PART-D|11 Videos

Similar Questions

Explore conceptually related problems

What is interference of light?

Explain the theory of interference of light.

Give any two characteristics of interference of light waves.

What is fringe width?