# Young's Double Slit Experiment

by cupid.callin
Tags: double, experiment, slit, young
 P: 1,135 1. The problem statement, all variables and given/known data so here i have shown the distances of two dark fringes ... from here ... fringe width is $$\frac{\lambda L}{2d}$$ L is distance of screen ... d is distance b/w 2 fringes. but fringe width is $$\frac{\lambda L}{d}$$ , right ??? What mistake am i making? Attached Thumbnails
 P: 463 What's the question?
 P: 1,135 i wrote it above by using formula of distance of dark fringe ... fringe width comes out to be ... $$\frac{\lambda L}{2d}$$ but fringe width is $$\frac{\lambda L}{d}$$ so why am i getting wrong answer
P: 1,135

## Young's Double Slit Experiment

hello ... someone?
 PF Patron P: 1,890 The fringe width is the distance between the centres of alternate dark(or bright) fringes. Lamda D/d as marked on your diagram goes from the centre of a bright fringe to the centre of an adjacent dark fringe in other words you have marked in half the fringe width.
 P: 1,135 So you mean this is fringe width ?? well this looks satisfying Attached Thumbnails
PF Patron
P: 1,890
 Quote by cupid.callin So you mean this is fringe width ?? well this looks satisfying
You have now marked the fringe width correctly.
 P: 1,135 are you sure? because fringe width means "width of 1 fringe" but (as you said) i have marked width of 2 fringes
 PF Patron P: 1,890 I did not say you marked in the width of two fringes I said you marked in the width of half a fringe.The fringe width as marked above in red is correct since it goes from the centre of a bright fringe to the centre of an adjacent bright fringe. Look at a graph of how the intensity varies smoothly across an interference pattern and this should clarify things for you.

 Related Discussions Introductory Physics Homework 2 General Physics 4 Classical Physics 5 Introductory Physics Homework 1 Classical Physics 3