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In a double slit experiment using light from a helium-neon laser, a student obtained the following results:

width of 10 fringes = 1.5 cm

separation of slits a = 1.0 mm

slit-to-screen distance D = 2.40 m

Determine the wavelength of light.

To solve the problem, I have to find the fringe separation. If I divide $1.5$ by $10$, I would get the width of $1$ fringe. Is that the same thing as the fringe separation?

A diagram showing what the fringe separation is would be really helpful. The textbook that I'm currently reading defines the fringe separation as the distance between the centres of adjacent bright fringes. What does that mean?

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closed as off-topic by John Rennie, Buzz, Jon Custer, M. Enns, ZeroTheHero Dec 17 '18 at 2:42

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  • $\begingroup$ Why are you asking for fringe separation when the question is asking for wavelength? $\endgroup$ – resgh Mar 3 '14 at 11:45
  • $\begingroup$ And yes, you're right. Go ahead. $\endgroup$ – resgh Mar 3 '14 at 11:48
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Double slit interference pattern looks as shown below:
enter image description here

When you see the same patter at different angle you will see it as shown below: enter image description here

Believing that your text book is correct in saying that fringe separation as the distance between the centers of adjacent bright or dark fringes (in double slit experiment) from the center of the screen, my textbook defines fringe width as the same (by the formula of fringe width). So, I think fringe width is nothing but fringe separation.

I think you are facing difficulty in understanding how dark fringe could have width seeing the second image above. I hope you are understanding bright fringe width. The dark fringe width can be assumed to be distance between upper points of the two consecutive crest like shapes (see the below figure, it is the wave, but you can imagine it to be interference pattern), it mean the same as saying distance between the centers of two consecutive bright or dark fringe.

enter image description here

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  • $\begingroup$ I'm still confused. $\endgroup$ – Asad Moosvi Mar 3 '14 at 16:23
  • $\begingroup$ @AsadMoosvi: I can try to help you, if you would say where you are confused. I have edited once again, please see the changes. $\endgroup$ – Immortal Player Mar 3 '14 at 17:47
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It is fringe width ,the distance between centers of adjacent secondary maxima's or minima's and it is $\lambda d/D$.This is what question expects from you here to use.But instead I think you are confused because actual width of fringe looks only half of above.But note it is varying intensity of light there so you choose only lines as fringe centers and just calculate distance between them to have fringe width.Just read this carefully to understand I neither explained it in detail nor explained it as good as I can.

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