Chapter 33, Problem 31P

(a)

To determine

To show: The condition for a constructive bright interference ring, thickness $\left(2t\right)$ is $(m+\frac{1}{2})\lambda$ .

Explanation of Solution

Introduction:

The given arrangement of Newton’s ring apparatus which consists of a plano-convex glass lens of radius of curvature R placed on a flat horizontal glass plate (Refer Figure 33-42) is similar to that of thin film arrangement with a difference that here the film is air of variable thickness $t$ .

Therefore, the phase change takes place at the top of the horizontal glass plate is $180°\left(\frac{1}{2}\lambda \right)$ .

Write the expression for the thickness of the thin glass plate in constructive interference.

  $2t+\frac{1}{2}\lambda =\lambda ,2\lambda ,3\lambda ,\mathrm{...}$

Here, $t$ is the thickness of the thin film and $\lambda$ is the wavelength of the light used.

Simplify the above expression and rearrange the terms for thickness.

  $\begin{array}{c}2t+\frac{1}{2}\lambda =\frac{1}{2}\lambda ,\frac{3}{2}\lambda ,\frac{5}{2}\lambda ,\mathrm{...}\\ 2t=\left(m+\frac{1}{2}\right)\lambda \end{array}$

where $m=0,1,2,\mathrm{...}$ .

Conclusion:

Thus, the condition for a constructive bright interference ring, thickness $\left(2t\right)$ is $(m+\frac{1}{2})\lambda$ .

(b)

To determine

To show: The relation between radius of the fringe $r$ and the thickness $t$ of the thin film is $\sqrt{2tR}$ for $t<<R$ , where $R$ is the radius of the curvature.

Explanation of Solution

Introduction:

The given arrangement of Newton’s ring apparatus which consists of a plano-convex glass lens of radius of curvature R placed on a flat horizontal glass plate (Refer Figure 33-42) is similar to that of thin film arrangement with a difference that here the film is air of variable thickness $t$ .

Therefore, considering the geometry of plano-convex glass lens and the horizontal glass plate on which the lens arrangement is placed.

Write the expression for the radius of curvature.

  $R^2={\left(R-t\right)}^2+r^2$

Here $R$ is the radius of curvature, $r$ is the radius of fringe and $t$ is the thickness of the thin film.

Simplify and expand the above expression as:

  $\begin{array}{l}{R}^2=r^2+\left(R^2-2Rt+t^2\right)\\ R^2=r^2+R^2-2Rt+t^2\end{array}$

When $t<<R$ , the term $t^2$ can be neglected as $t$ on squaring becomes very small in comparison of $R$ . Therefore,

Write again the above expression forradius of curvature.

  $R^2\approx r^2+R^2-2Rt$

Simplify the above expression forthe radius of fringe.

  $\begin{array}{c}{r}^2=R^2-R^2+2Rt\\ r^2=2Rt\\ r=\sqrt{2Rt}\end{array}$ ……(1)

Conclusion: Thus, for $t<<R$ the radius of the fringe $r$ and the thickness $t$ of the thin film is related as $\sqrt{2tR}$ .

(c)

To determine

The number of bright fringes obtained in the reflected light.

Explanation of Solution

Given:

The radius of curvature of the plano-convex lensis $10\text{m}$ .

The diameter of the lensis $4\text{cm}$ .

The wavelength of the light used is $590\text{nm}$ .

Formula used:

Write the expression for the radius of curvature.

  $R^2={\left(R-t\right)}^2+r^2$

Here $R$ is the radius of curvature, $r$ is the radius of fringe and $t$ is the thickness of the thin film.

Simplify and expand the above expression as:

  $\begin{array}{l}{R}^2=r^2+\left(R^2-2Rt+t^2\right)\\ R^2=r^2+R^2-2Rt+t^2\end{array}$

When $t<<R$ , the term $t^2$ can be neglected as $t$ on squaring becomes very small in comparison of $R$ . Therefore,

Write again the above expression forradius of curvature.

  $R^2\approx r^2+R^2-2Rt$

Simplify the above expression forthe radius of fringe.

  $\begin{array}{c}{r}^2=R^2-R^2+2Rt\\ r^2=2Rt\end{array}$ …… (2)

Write the expression for the thickness of the thin glass plate in constructive interference.

  $2t+\frac{1}{2}\lambda =\lambda ,2\lambda ,3\lambda ,\mathrm{...}$

Here $t$ is the thickness of the thin film and $\lambda$ is the wavelength of the light used.

Simplify the above expression and rearrange the terms for thickness.

  $\begin{array}{c}2t+\frac{1}{2}\lambda =\frac{1}{2}\lambda ,\frac{3}{2}\lambda ,\frac{5}{2}\lambda ,\mathrm{...}\\ 2t=\left(m+\frac{1}{2}\right)\lambda \\ t=\left(m+\frac{1}{2}\right)\frac{\lambda }{2}\end{array}$

where $m=0,1,2,\mathrm{...}$ .

Substitute $\left(m+\frac{1}{2}\right)\frac{\lambda }{2}$ for $t$ in equation (1).

  $\begin{array}{c}{r}^2=2\left(m+\frac{1}{2}\right)\frac{\lambda }{2}R\\ =\left(m+\frac{1}{2}\right)R\lambda \end{array}$

Rearrange the above expression for the number of fringes.

  $\begin{array}{c}m+\frac{1}{2}=\frac{r^2}{R\lambda }\\ m=\frac{r^2}{R\lambda }-\frac{1}{2}\end{array}$ ......(3)

Calculation:

Substitute $2\text{cm}$ for $r$ , $10\text{m}$ for $R$ and $590\text{nm}$ for $\lambda$ in equation (3).

  $\begin{array}{c}m=\frac{{\left(2\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}^2}{\left(10\text{m}\right)\left(590\text{nm}\left(\frac{1\text{m}}{{10}^9\text{nm}}\right)\right)}-\frac{1}{2}\\ =67.2\\ \approx 68\end{array}$

Conclusion:

Thus, there are $68$ bright fringes obtained in the reflected light.

(d)

To determine

The diameter of the sixth bright fringe.

Explanation of Solution

Given:

The value of $m$ for sixth bright fringe is $5$ .

The radius of curvature of the plano-convex lensis $10\text{m}$ .

The wavelength of the light used is $590\text{nm}$ .

Formula used:

Write the expression for the radius of curvature.

  $R^2={\left(R-t\right)}^2+r^2$

Here $R$ is the radius of curvature, $r$ is the radius of fringe and $t$ is the thickness of the thin film.

Simplify and expand the above expression as:

  $\begin{array}{l}{R}^2=r^2+\left(R^2-2Rt+t^2\right)\\ R^2=r^2+R^2-2Rt+t^2\end{array}$

When $t<<R$ , the term $t^2$ can be neglected as $t$ on squaring becomes very small in comparison of $R$ . Therefore,

Write again the above expression forradius of curvature.

  $R^2\approx r^2+R^2-2Rt$

Simplify the above expression forthe radius of fringe.

  $\begin{array}{c}{r}^2=R^2-R^2+2Rt\\ r^2=2Rt\\ r=\sqrt{2Rt}\end{array}$

Write the expression for the thickness of the thin glass plate in constructive interference.

  $2t+\frac{1}{2}\lambda =\lambda ,2\lambda ,3\lambda ,\mathrm{...}$

Here $t$ is the thickness of the thin film and $\lambda$ is the wavelength of the light used.

Simplify the above expression and rearrange the terms for thickness.

  $\begin{array}{c}2t+\frac{1}{2}\lambda =\frac{1}{2}\lambda ,\frac{3}{2}\lambda ,\frac{5}{2}\lambda ,\mathrm{...}\\ 2t=\left(m+\frac{1}{2}\right)\lambda \\ t=\left(m+\frac{1}{2}\right)\frac{\lambda }{2}\end{array}$

where $m=0,1,2,\mathrm{...}$ .

Write the expression for the diameter $m^{th}$ bright fringe.

  $D=2r$

Substitute $\sqrt{2Rt}$ for $r$ in the above expression.

  $D=2\sqrt{2Rt}$

Substitute $\left(m+\frac{1}{2}\right)\frac{\lambda }{2}$ for $t$ in the above expression.

  $\begin{array}{c}D=2\sqrt{2R\left(m+\frac{1}{2}\right)\frac{\lambda }{2}}\\ =2\sqrt{\left(m+\frac{1}{2}\right)R\lambda }\end{array}$ …….(4)

Calculation:

Substitute $5$ for $m$ , $10\text{m}$ for $R$ and $590\text{nm}$ for $\lambda$ in equation (4).

  $\begin{array}{c}D=2\sqrt{\left(5+\frac{1}{2}\right)\left(10\text{m}\left(\frac{{10}^2\text{cm}}{1\text{m}}\right)\right)\left(590\text{nm}\left(\frac{1\text{cm}}{{10}^{-7}\text{nm}}\right)\right)}\\ =1.14\text{cm}\end{array}$

Conclusion:

Thus, the diameter of the sixth bright fringe is $D=1.14\text{cm}$ .

(e)

To determine

The qualitative changes occur in the bright fringe pattern when air is replaced by the water between the two glass plates.

Explanation of Solution

Given:

The wavelength of the light used is $590\text{nm}$ .

The refractive index of the glass plates, $n$ is $1.5$

Formula used:

Write the expression for wavelengthof the light in the film.

  $\frac{{\lambda }_{\text{air}}}{n}=444\text{nm}$

Rearrange the above expression for refractive index.

  $n=\frac{{\lambda }_{\text{air}}}{444\text{nm}}$ ……(6)

Calculation:

Substitute $590\text{nm}$ for ${\lambda }_{\text{air}}$ in equation (6).

  $\begin{array}{c}n=\left(\frac{590\text{nm}}{444\text{nm}}\right)\\ =1.33\end{array}$

Conclusion:

Thus, the number of bright fringe pattern increased by a factor of $1.33$ .