Question

A Newton’s rings apparatus is to be used to determine the radius of curvature of a lens . The radii of the nth and $\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{20}\mathbf{th}\mathbf{)}$bright rings are found to be $\mathbf{0}\mathbf{.}\mathbf{162}\mathbf{}\mathbf{cm}$ and $\mathbf{0}\mathbf{.}\mathbf{368}\mathbf{}\mathbf{cm}$, respectively, in light of wavelength $\mathbf{546}\mathbf{}\mathit{n}\mathit{m}$. Calculate the radius of curvature of the lower surface of the lens.

Short answer

The radius of the curvature of the lower surface of the lens is $1.0\mathrm{m}$.

Step-by-step solution

Given data

The radii of$nth=0.162\text{cm}$

The radii of $\left(n+20\right)=0.368\text{cm}$

Wavelength $\lambda =546\text{nm}$

Definition of newton’s ring

Newton's ring is a phenomenon in which an interference pattern is created by the reflection of light between two surfaces; a spherical surface and an adjacent touching flat surface.

Determine the radius of the curvature 

The radius of curvature of the xth ring

$R_x=0.162\text{cm}$

For ${\left(n+20\right)}^{th}{R}_{n+20}=0.368\text{cm}$

Let the radius of the curvature of the lower surface of the lens be R.

The value of R is given by as

$\begin{array}{rcl}R& =& \frac{R_{n+20}^2-R_x^2}{20\times \lambda }\\ R& =& \frac{{\left(0.368\times {10}^{-2} \text{m}\right)}^2-{\left(0.162\times {10}^{-2} \text{m}\right)}^2}{20\times 546\times {10}^{-9} \text{m}}\\ R& =& 0.9998\text{m}\\ R& =& 1.0\text{m}\end{array}$

Hence, the radius of the curvature of the lower surface of the lens is $1.0\mathrm{m}$.