Answer to Question #177670 in Mechanical Engineering for aryan

Question

The diameter of the 9th dark ring in the newton’s ring experiment is 0.28 cm. What is the diameter of

16th dark ring? Given the wavelength of light used is 6000 A0

. Calculate the radius of curvature of the

lens used.

Answer 1

$D_n = 0.28 cm = 0.0028 m$

$\lambda = 6000 A^0 = 6000 \times 10^{-10}m$

Radius of curvature

$R = \frac {D^2_n}{4n \lambda}$

$R = \frac {0.0028^2}{4 \times 9 \times 6000 \times 10^{-10}} = 0.3629 m = 36.29 cm$

 diameter of 16th dark ring

$D_{16} = \sqrt{R \times 4 n \lambda}$

$D_{16} = \sqrt{0.3629 \times 4\times16 \times 6000 \times 10^{-10}} = 0.003733 m =0.3733 cm$