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A Newton's rings set up is used with a source emitting two wavelength $\lambda_{1}$=6000A$^{\circ}$ and $\lambda_{2}$=4500A$^{\circ}$.

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°.

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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 …

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