Question

In Young's double slit experiment the distance between the slits S1 & S2is 1mm. What should the width of each slit be so as to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern??? Explain it clearly nd ans soon.....

Answer 1

You are confusing interference with diffraction. Distance between fringes depends on the distance between slits only. So, when you keep the slit distance constant and change their widths, the distance between fringes stays constant, and the number of them visible depends on the size of the part of the screen that is lit. And this increases with increasing diffraction, i.e. with decreasing slit width. Oh, and the slit width can't really be "much wider than the wavelength of light", because the distance of them wouldn't be comparable to the wavelength, and you'd get no interference pattern at all.

HOPE IT IS HELPFUL.

Answer 2

Let a be the width of each slit. Linear separation between 10 bright fringes.

x = 10β =

d

10λD

corresponding angular separation

Θ

1

= x/D = 10λ/d

Now, the angular width of central maximum in the diffraction pattern of a single slit,

Θ

2

=2λ/a

As Θ

2

=Θ

1

2λ/a=10λ/d

or a = d/5 = 1.00/5 mm = 0.2mm