Question

A curved piece of glass with a radius of curvature R rests on a flat plate of glass. Light of wavelength ? is incident normally on this system.

Considering only interference between waves reflected from the curved (lower) surface of glass and the top surface of the plate, find the radius of the nth dark ring.

Answer 1

there are two surfaces one is curved surface and the other is the plane surface of the glass

   when the light falls on the surfaces and reflect there will be a phase change in the reflected light

   depending upon the surface

   for the curved surface we get

   phase change = 2 n_glass t / ?and

   for the plate we get

   phase change = (2 n_glass t / ?) +(2 d / ?) + (1 / 2)

   where t is the thickness of the curved glass andd is the distance from the bottom of the curved

   glass to the top of the plate

   the difference in the phase changes will be = (2n_glass t / ?) + (2 d / ?) + (1 / 2) - (2n_glass t / ?)

                                                                  = (2 d / ?) + (1 / 2)

   as the radius of the nth dark ring is required,the dark rings occurs when

   (2 d / ?) + (1 / 2) = (1 / 2), (3 / 2), (5 /2), ..............

   (2 d / ?) = 0, 1, 2,...................

                =n

   n = 0, 1, 2, .................

   so in nth terms we can write

   d_n = n ? / 2

   let us assume that the distance from then^th dark ring be one leg of the right angled triangle with

   radius of curvature as the hypotenuse, thedistance from the center of curvature to the point of

   contact of the two pieces of glass equals theradius of curvature, this distance minus dn will be

   the other leg if the triangle

   so we can use the pythagorean theorm to get rnas

   R² = r_n² + (R -d_n)²

   r_n² = R² -R² + 2 R d_n - d_n²

   r_n = ?2 R d_n -d_n²

       = ? 2 R (n? / 2) - (n ? / 2)²

       = n ? R -(n² ?² / 4)