It is found that nth dark ring due to 6000A° coincides with (n+2)th dark ring due to 4500A°. If the radius of curvature of the lens is 90 cm, find the diameter of nth dark ring for 6000A°.

Solution:

$ \lambda_1 = 6000 A^0 = 6000 \times 10^{-8} cm $ for $D_n^2$

$ \lambda_2 = 4500 A^0 = 4500 \times 10^{-8} cm $ for $D_{n+1}^2$

$R = 90 cm$

$\mu = 1$

Find $r_n = ?$

$\begin{aligned} D_n^2 &= 4 Rn \lambda \\ D_n^2 &= 4 Rn \lambda_1 \\ D_{n+1}^2 &= 4 R(n+1) \lambda_2 \\ \end{aligned}$

$\begin{aligned} D_n^2 &= D_{n+1}^2 \\ 4Rn \lambda_1 &= 4 R(n+1) \lambda_2 \\ n &= \frac{\lambda_1}{\lambda_1 - \lambda_2} \\ &= \frac{4500 \times 10^{-8}}{(6000 - 4500) \times 10^{-8}} \\ &= 3 \\ \end{aligned}$

$\begin{aligned} D_n^2 &= 4 \times 90 \times 3 \times 6000 \times 10^{-8} \\ &= 648 \times 10^{-4} \end{aligned}$

$\begin{aligned} D_n &= 25.45 \times 10^{-2} \ cm\\ &= 2545 \ cm \\ r_n &= 0.504 \ cm \end{aligned}$

$\therefore \text{ Radius of } n^{th} \text{dark ring = 0.504 cm}$