Given :

$D_5 = 0.336 cm$

$D_15 = 0.590 cm$

$λ = 5890 A^o = 5.89 × 10^{-5} cm$

To find: R

Solution:

$R = \frac{D_{15}^2 - D_5^2}{4λ (15 - 5)} = 99.83cm$

A planoconvex lens of large radius of curvature is placed on a plane glass plate so that a thin film of air is trapped between the spherical surface of the lens and a glass plate. This thin air film is illuminated by a monochromatic source of light. The light from the source is rendered parallel by means of a converging lens. The parallel beam of light is intercepted by means of a glass plate inclined at an angle of 45° so as to get the normal incidence. This light illuminates the air film and the interference pattern is observed through a traveling microscope focussed on it.

The interference pattern consists of dark and bright concentric rings with dark centre when observed in reflected light. The cross wire of the eyepiece of an traveling microscope is focussed on $n^{th}$ dark ring and microscopic readings(main scale reading and Vernier scale readings) are noted.

$D_n^ 2$ = 4nλR / μ -------------------------------(1)

The traveling microscope is moved and focussed on (n+p) th dark ring,and again microscopic readings are noted.

$D_{(n+p)}^2$ = 4(n+p)λR /μ-------------------------------(2)

From equation (1) and (2)

$D_{(n+p)}^2$ - $D_n^2$ = 4pλR / μ

Diameters can be found experimentally , knowing the wavelength of the source ,radius of curvature can be determined.