Question

Show that the fringe width decreases with the order of rings​

Answer 1

Given:

Fringe width.

To Find:

Show that the fringe width decreases with the order of rings?

Solution:

firstly, let discuss about fringe-

• Fringe width is the distance between two successive bright fringes or two successive dark fringes.

• Fringe width is independent of order of fringe. Fringe width is directly proportional to wavelength of the light used.

It is given by:

$\[\beta = \dfrac{{\lambda D}}{d}\]$      

where, d is the distance between the slits , λ is the wavelength and D is the distance between source and screen .

Also, we know that-

Rings are not equally spaced because the diameter of ring does not increase in the same proportion as the order of ring and rings get closer and closer as ‘n’ increases.

as, diameter of dark ring is given by $D_{n}=\sqrt{4n\lambda R}$  where, n=0,1,2,3,..

eg.

$D_{5} -D_{4}=(\sqrt{5}-\sqrt{4} )\sqrt{\lambda R} \\=0.236\sqrt{\lambda R}$

As, the order of rings (n) increases,  D decreases.

Hence, the fringe width decreases with the order of rings.