Q36P - University Physics with Modern Physics

Question

Newton’s rings are visible when a plano-convex lens is placed on a flat glass surface. For a particular lens with an index of refraction of n = 1.50 and a glass plate with an index of n = 1.80, the diameter of the third bright ring is 0.640 mm. If water 1n = 1.332 now fills the space between the lens and the glass plate, what is the new diameter of this ring? Assume the radius of curvature of the lens is much greater than the wavelength of the light.

Step-by-Step Solution

Answer

0.555 nm

Step 1: Given

$n_{\text{lenz }}=1.50,n_{\text{plate }}=1.80,n_{\text{air }}=1.0,m=3.0,D_3=0.640\mathrm{mm}=0.640\times {10}^{-3}\mathrm{m},n_{\text{water }}=1.33$

Step 2: Basic concept.

In the first case, when the third ring has a diameter of 0640 mm, the thein film between the glass plate and the convex lens is the air ($n_{air}$ =1.0)

As you see below, the refracted ray from the upper surface of the air film experience no phase change (since the index of refraction of the lens is greater than that of the ain), but the second ray experiences a phase change (since the index of refraction of the air film is less than that of the plate) 

The red circle indicates a phase change $180°$.

Noting that, when the air is replaced by the water, we will still have the same phase change due to the second ray reflection since

$n_{\text{lenz }}>n_{\text{water }}\text{ And }{n}_{\text{water }}<n_{\text{plate }}$  

From all the above, we have two reflected rays with one phase change. So, the thickness of this bright fringe for this case is given by

                                   $\begin{array}{r}2\mathrm{t}=\left(m+\frac{1}{2}\right){\lambda }_{\text{film }}\\ t=\frac{\left(m+\frac{1}{2}\right){\lambda }_{\text{flm }}}{2}\end{array}$                                                                 (1)

                   

Now we need to find the radius of the third bright fringe in terms of t. 

We will magnify the figure above, so we can find , as you see below.

Step 3: Ray diagram.

                                 

                          

                 

Step 4: Calculation.

We analyzed the whole given in the figure above. Now the problem is much like a geometrical problem. 

From the Right-blue-triangle in the last figure above, and by applying the Pythagorean theorem, we can find $r_3$ which is given by

                                           $$\begin{gathered} {\mathrm{R}}^2=(\mathrm{R}-\mathrm{t}{)}^2+{\mathrm{r}}_3^2 \\ \mathrm{Hence}, \\ \begin{array}{c}{\mathrm{r}}_3^2={\mathrm{R}}^2-(\mathrm{R}-\mathrm{t}{)}^2\\ {\mathrm{r}}_3^2={\mathrm{R}}^2-\left({\mathrm{R}}^2-2\mathrm{Rt}+{\mathrm{t}}^2\right)\\ {\mathrm{r}}_3^2={\mathrm{R}}^2-{\mathrm{R}}^2+2\mathrm{Rt}-{\mathrm{t}}^2\\ {\mathrm{R}}^2-{\mathrm{R}}^2=0\\ {\mathrm{r}}_3^2=2\mathrm{Rt}-{\mathrm{t}}^2\\ \mathrm{Hence}, \end{array} \\ {\mathrm{r}}_3=\sqrt{2\mathrm{Rt}-{\mathrm{t}}^2} \end{gathered}$$  

substitute t from (1);

                                           $r_3=\sqrt{\left[2R\times \frac{\left(m+\frac{1}{2}\right){\lambda }_{\text{film }}}{2}\right]-{\left[\frac{\left(m+\frac{1}{2}\right){\lambda }_{\text{film }}}{2}\right]}^2}$

Noting that the author told us to assume that the radius of the curvature R of the lens is much greater than the wavelength of the light 

Also, note that the wavelength of the light is for power  which means when you square it the value will be too small to be counted. 

So that the term ${\left[\frac{\left(m+\frac{1}{2}\right){\lambda }_{\text{flm }}}{2}\right]}^2$ is close to zero. 

Hence,

                   $\begin{array}{r}{r}_3=\sqrt{\left[2R\times \frac{\left(m+\frac{1}{2}\right){\lambda }_{\text{film }}}{2}\right]}-0\\ \left.r_3\right]\sqrt{2R\times \frac{\left(m+\frac{1}{2}\right){\lambda }_{\text{film }}}{2}}\\ r_3=\sqrt{\left(m+\frac{1}{2}\right){\lambda }_{\text{film }}R}\end{array}$

                                                                                          ………..(2)

For the first case, when the film between the lens and the plate is air, equation (2) will be as follows.

                                $r_3=\sqrt{\left(m+\frac{1}{2}\right){\lambda }_{air}R}$

Noting that $r_3=\frac{D_3}{2}$ hence,

                                    $\frac{D_3}{2}=\sqrt{\left(m+\frac{1}{2}\right){\lambda }_{\text{air }}R}$                   

solving for ${\lambda }_{air}$ R, so we will square both sides.

                           $\begin{array}{c}{\left(\frac{D_3}{2}\right)}^2=\left(m+\frac{1}{2}\right){\lambda }_{\text{air }}R\\ \frac{D_3^2}{4\left(m+\frac{1}{2}\right)}={\lambda }_{\text{air }}R\\ {\lambda }_{\text{air }}R=\frac{D_3^2}{4\left(m+\frac{1}{2}\right)}\\ {\lambda }_{\text{air }}R=\frac{{\left(0.640\times {10}^{-3}\right)}^2}{4\left(3+\frac{1}{2}\right)}\\ {\lambda }_{\text{air }}R=2.93\times {10}^{-8}m\end{array}$

                                                                                              (3)

When the film between the lens and the plate is water, equation (2) will be as follows.

                                 $r_{3,\text{ new }}=\sqrt{\left(m+\frac{1}{2}\right){\lambda }_{\text{water }}R}$                                          (4)

We need to find the wavelength of the light inside the ater in terms of  We know, from Snell's law, that

                                          $n_1{\lambda }_1=n_2{\lambda }_2$                       

Hence,

                                            $n_{\text{flim }}{\lambda }_{\text{flim }}=n_{\text{water }}{\lambda }_{\text{water }}$           

solving for ${\lambda }_{\text{water }}$ and noting that $n_{\text{water }}$ = 1.0;

                                              ${\lambda }_{\text{water }}=\frac{{\lambda }_{\text{air }}}{n_{\text{water }}}$                 

Substitute into (4)

                               $\begin{array}{c}{r}_{3\text{ new }}=\sqrt{\left(m+\frac{1}{2}\right)\times \frac{{\lambda }_{\text{air }}}{n_{\text{water }}}\times R}\\ r_{3\text{, new }}=\sqrt{\left(m+\frac{1}{2}\right)\times \frac{{\lambda }_{\text{air }}R}{n_{\text{water }}}}\end{array}$

Substitute ${\lambda }_{air}$ R from (3) and Substitute the given.

                                $\begin{array}{c}{r}_{3,\text{ new }}=\sqrt{\left(3+\frac{1}{2}\right)\times \frac{2.93\times {10}^{-8}\mathrm{m}}{1.33}}\\ r_{3,\text{ new }}=2.77\times {10}^{-4}\mathrm{m}\end{array}$                    

Now we can easily find the diameter of the third ring, which is given by

                                   $\begin{array}{r}D{r}_{3,\text{ new }}=2r_{3,\text{ new }}\\ D{r}_{3,\text{ new }}=2\times 2.77\times {10}^{-4}\mathrm{m}\\ D{r}_{3,\text{ new }}=5.55\times {10}^{-4}\mathrm{m}\end{array}$