Question

In a Young's double slit experiment, the distance between the slits is 0.2 mm, the slits are at a distance 2 m from the screen, and monochromatic light of wavelength 500 nm is used. The distance between two closest points on the screen where intensity is one-fourth the maximum intensity on the screen is mm.​

Answer 1

Answer:

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