Chapter 33, Problem 36P

(a)

To determine

The separation of the slits.

Explanation of Solution

Given:

The wavelength of the light used is $633\text{nm}$ .

The distance between the slit and screen is $12\text{m}$ .

The distance of first interference from the central maxima is $82\text{cm}$ .

Formula used:

The figure below shows the geometry of the given setup.

  

Here $L$ is the distance between the slit and screen, $y_1$ is the distance of first interference from the central maxima, $\lambda$ is the wavelength of the light used, $d$ is the distance between the two slits, ${\theta }_1$ is the angle between first interference and the central maxima and $\theta$ is the path difference between the two waves.

Write the expression for two-slit interference maxima for $d<<L$ .

  $\sin {\theta }_1\approx \frac{\lambda }{d}$

Rearrange the above expression for the distance between the two slits.

  $d\approx \frac{\lambda }{\sin {\theta }_1}$   ...... (1)

Write the expression for the triangle in the above figure with sides $L$ and $y_1$ .

  $\sin {\theta }_1=\frac{y_1}{\sqrt{{\left(L\right)}^2+{\left(y_1\right)}^2}}$   ...... (2)

Calculation:

Substitute $12\text{m}$ for $L$ and $82\text{cm}$ for $y_{\text{1}}$ in equation (2).

  $\begin{array}{c}\sin {\theta }_1=\frac{\left(82\text{cm}\right)}{\sqrt{{\left(12\text{m}\left(\frac{{10}^2\text{cm}}{\text{m}}\right)\right)}^2+{\left(82\text{cm}\right)}^2}}\\ =\frac{\left(82\text{cm}\right)}{(1202.8\text{cm})}\\ =0.06817\end{array}$

Substitute $633\text{nm}$ for $\lambda$ and $0.06817$ for $\sin {\theta }_1$ in equation (1).

  $\begin{array}{c}d\approx \frac{\left(633\text{nm}\right)}{0.06817}\\ =9285.6\text{nm}\\ \text{=}9285.6\text{nm}\left(\frac{\text{1}\text{μm}}{{10}^3\text{nm}}\right)\\ \approx 9.3\text{μm}\end{array}$

Conclusion:

Thus, the separation of the slits is $9.3\text{μm}$ .

(b)

To determine

The possible number of interference fringes.

Explanation of Solution

Given:

The wavelength of the light used is $633\text{nm}$ .

The distance between the slit and screen is $12\text{m}$ .

The distance of first interference from the central maxima is $82\text{cm}$ .

Formula used:

Write the expression for two-slit interference maxima for $d<<L$ .

  $\sin {\theta }_1\approx \frac{\lambda }{d}$

Here $L$ is the distance between the slit and screen, $\lambda$ is the wavelength of the light used, ${\theta }_1$ is the angle between first interference and $d$ is the distance between the two slits.

Rearrange the above expression for the distance between the two slits.

  $d\approx \frac{\lambda }{\sin {\theta }_1}$   ...... (1)

Write the expression for the triangle in the above figure with sides $L$ and $y_1$ .

  $\sin {\theta }_1=\frac{y_1}{\sqrt{{\left(L\right)}^2+{\left(y_1\right)}^2}}$   ...... (2)

Here $y_1$ is the distance of first interference from the central maxima.

Write the expression for the two slit interference maxima.

  $d\sin \theta =m\lambda$

where $m$ is an integer.

Write the above expression for maximum number of interference fringes, for $\sin \theta \le 1$ .

  $d=m_{\text{max}}\lambda$

Here $m_{\text{max}}$ is the maximum bright fringe.

Simplify above expression for $m_{\text{max}}$ .

  $m_{\text{max}}=\frac{d}{\lambda }$   ...... (3)

Write the expression for the maximum number of fringes in the central maximum.

  $N=2m_{\text{max}}+1$   ...... (4)

Here $N$ is the number of fringes in the central maximum

Calculation:

Substitute $12\text{m}$ for $L$ and $82\text{cm}$ for $y_{\text{1}}$ in equation (2).

  $\begin{array}{c}\sin {\theta }_1=\frac{\left(82\text{cm}\right)}{\sqrt{{\left(12\text{m}\left(\frac{{10}^2\text{cm}}{\text{m}}\right)\right)}^2+{\left(82\text{cm}\right)}^2}}\\ =\frac{\left(82\text{cm}\right)}{(1202.8\text{cm})}\\ =0.06817\end{array}$

Substitute $633\text{nm}$ for $\lambda$ and $0.06817$ for $\sin {\theta }_1$ in equation (1).

  $\begin{array}{c}d\approx \frac{\left(633\text{nm}\right)}{0.06817}\\ =9285.6\text{nm}\\ \text{=}9285.6\text{nm}\left(\frac{\text{1}\text{μm}}{{10}^3\text{nm}}\right)\\ \approx 9.3\text{μm}\end{array}$

Substitute $9.3\text{μm}$ for $d$ and $633\text{nm}$ for $\lambda$ in equation (3).

  $\begin{array}{c}{m}_{\text{max}}=\frac{\left(9.3\text{μm}\right)\left(\frac{{10}^3\text{nm}}{1\text{μm}}\right)}{633\text{nm}}\\ =\frac{9300}{633}\\ =14\end{array}$

Substitute $14$ for $m_{\text{max}}$ in equation (4).

  $\begin{array}{c}N=2\left(14\right)+1\\ =28+1\\ =29\end{array}$

Conclusion:

Thus, the possible number of interference fringes is $29$ .