Chapter 38, Problem 12P

Coherent light of wavelength 501.5 nm is sent through two parallel slits in an opaque material. Each slit is 0.700 μm wide. Their centers are 2.80 μm apart. The light then falls on a semicylindrical screen, with its axis at the midline between the slits. We would like to describe the appearance of the pattern of light visible on the screen. (a) Find the direction for each two-slit interference maximum on the screen as an angle away from the bisector of the line joining the slits. (b) How many angles are there that represent two-slit interference maxima? (c) Find the direction for each single-slit interference minimum on the screen as an angle away from the bisector of the line joining the slits. (d) How many angles are there that represent single-slit interference minima? (e) How many of the angles in part (d) are identical to those in part (a)? (f) How many bright fringes are visible on the screen? (g) If the intensity of the central fringe is I_max, what is the intensity of the last fringe visible on the screen?

To determine

The direction for the each two slit interference as an angle away from the bisector of the line joining the centre of the slits.

Answer to Problem 12P

The possible direction of two slit interference maxima are $0°$, $\pm 10.3°$, $\pm 20.9°$, $\pm 32.4°$, $\pm 45.7°$ and $\pm 63.5°$ angle away from the bisector of the line joining the slits..

Explanation of Solution

Given information: The wavelength of light is $501.5\text{nm}$, the width of each slit is $0.700\text{μm}$ and the distance between the centers of each slit is $2.80\text{μm}$.

The condition for double slit interference maxima is,

$d\sin \theta =m\lambda$ (1)

Here

$d$ is distance between the centers of each slit

$m$ is the order of interference

$\lambda$ is the wavelength of light

$\theta$ is the angle of interference

Further solve equation (1) as;

$\begin{array}{c}d\sin \theta =m\lambda \\ \theta ={\sin }^{-1}\frac{m\lambda }{d}\end{array}$

Substitute $501.5\text{nm}$ for $\lambda$ and $\text{2}\text{.80μm}$ for $\text{d}$ in the above equation.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{m\times 501.5\text{nm}}{2.80\mu m}\right)\\ ={\sin }^{-1}\left(\frac{m\times 501.5\text{nm}\times \left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{2.80\mu \text{m}\times \left(\frac{1\text{m}}{{10}^6\mu \text{m}}\right)}\right)\\ ={\sin }^{-1}\left(m\times 0.179\right)\end{array}$ (1)

For different value of $m$ as $m=0,1,2,3,\mathrm{...}$ calculate values of $\theta$ to obtain the directions of interference maxima.

For $m=0$:

Substitute $0$ for $m$ in equation (1).

$\begin{array}{c}{\theta }_0={\sin }^{-1}\left(0\times .179\right)\\ ={\sin }^{-1}\left(0\right)\\ =0\end{array}$

Thus for $m=0$, the direction of the zero order maxima is observed at $0°$ angle away from the bisector of the line joining the slits.

For $m=1$,

Substitute $1$ for $m$ in equation (1).

$\begin{array}{c}\theta ={\sin }^{-1}\left(1\times 0.179\right)\\ =\pm 10.3°\end{array}$

Thus for $m=1$, the direction of the first order maxima is observed at $\pm 10.3°$ angle away from the bisector of the line joining the slits.

For $m=2$,

Substitute $2$ for $m$ in equation (1).

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(2\times 0.179\right)\\ ={\sin }^{-1}\left(0.3582\right)\\ =\pm 20.9°\end{array}$

Thus for $m=2$, the direction of the second order maxima is observed at $\pm 20.9°$ angle away from the bisector of the line joining the slits.

For $m=3$:

Substitute $3$ for $m$ in equation (1).

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(3\times 0.179\right)\\ ={\sin }^{-1}\left(0.537\right)\\ =\pm 32.4°\end{array}$

Thus for $m=3$, the direction of the third order maxima is observed at $\pm 32.4°$ angle away from the bisector of the line joining the slits.

For $m=4$:

Substitute $4$ for $m$ in equation (1).

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(4\times 0.179\right)\\ ={\sin }^{-1}\left(0.716\right)\\ =\pm 45.7°\end{array}$

Thus for $m=4$, the direction of the fourth order maxima is observed at $\pm 45.7°$ angle away from the bisector of the line joining the slits.

For $m=5$:

Substitute $5$ for $m$ in equation (1).

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(5\times 0.179\right)\\ ={\sin }^{-1}\left(0.895\right)\\ =\pm 63.5°\end{array}$

Thus for $m=5$, the direction of the fifth order maxima is observed at $\pm 63.5°$ angle away from the bisector of the line joining the slits

For $m=6$:

Substitute $5$ for $m$ in equation (1).

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(6\times 0.179\right)\\ ={\sin }^{-1}\left(1.074\right)\end{array}$

The value of ${\sin }^{-1}\left(1.074\right)$ is not possible. Thus the diffraction maxima are possible up to fifth order only.

Conclusion:

Therefore, there are $11$ possible direction of two slit interference maxima are $0°$, $\pm 10.3°$, $\pm 20.9°$, $\pm 32.4°$, $\pm 45.7°$ and $\pm 63.5°$ angle away from the bisector of the line joining the slits.

To determine

The numbers of angles that represents two slit interference maxima.

Answer to Problem 12P

There are $11$ angles that represent two slit interference maxima.

Explanation of Solution

The calculation in part (a) shows that for zero order there is one angle and for first, second, third, fourth and fifth order there are each two direction for a single that represent the two slit interference maxima

The possible angles are $0°$, $\pm 10.3°$, $\pm 20.9°$, $\pm 32.4°$, $\pm 45.7°$ and $\pm 63.5°$. So there are total $11$ angles that represent two slit interference maxima.

Conclusion:

Therefore, there are $11$ angles of interference maxima.

To determine

The direction of each single-slit interference minimum on the screen as an angle away from the bisector of the line joining the slits.

Answer to Problem 12P

The direction of each single-slit interference minimum on the screen as an angle away from the bisector of the line joining the slits is.

Explanation of Solution

Given Info: The condition for the interference minima in single slit interference minima is,

$a\sin \theta =m\lambda$

Here,

$a$ is the slit width.

Substitute $0.700\mu m$ for $a$ and $501.5\text{nm}$ for $\lambda$ in the above equation.

$\begin{array}{c}0.700\text{μm}\left(\sin \theta \right)=m\left(501.5\text{μm}\right)\\ \theta ={\sin }^{-1}\left(\left(m\right)\frac{501.5\text{nm}\times \left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}{0.700\text{μm}\times \left(\frac{1\text{m}}{{10}^6\text{μm}}\right)}\right)\\ ={\sin }^{-1}\left(\left(m\right)0.716\right)\end{array}$ (2)

For different value of $m$ as $m=1,2,3,\mathrm{...}$ calculate values of $\theta$ to obtain the directions of interference minima.

For $m=1$:

Substitute $1$ or $m$ in the equation (2).

$\begin{array}{c}\theta ={\sin }^{-1}\left(\left(m\right)0.716\right)\\ ={\sin }^{-1}\left(\left(1\right)0.716\right)\\ =\pm 45.7°\end{array}$

Thus for $m=1$, the direction of the zero order single slit interference minima is observed at $\pm 45.7°$ angle away from the bisector of the line joining the slits.

For $m=2$:

Substitute $2$ or $m$ in equation (2).

$\begin{array}{c}\theta ={\sin }^{-1}\left(\left(m\right)0.716\right)\\ ={\sin }^{-1}\left(\left(2\right)0.716\right)\\ ={\sin }^{-1}\left(1.432\right)\end{array}$

The value of ${\sin }^{-1}\left(1.432\right)$ is not possible.

Thus up to second order the single slit interference is possible.

Conclusion:

Therefore, the possible directions are $\pm 45.7°$ for single slit interference minima away from the bisector of the line joining the slits.

To determine

The numbers of angles that represents single slit interference maxima.

Answer to Problem 12P

There are $2$ angles that represent two slit interference maxima.

Explanation of Solution

The calculation in part (c) shows that for first order only the single slit interference minima is possible.

The possible angles are $\pm 45.7°$.

So there are total $2$ angles that represent single slit interference minima.

Conclusion:

Therefore, there are $2$ angles represent single slit interference minima.

To determine

The numbers of angles that are identical for single interference minima and double slit interference maxima.

Answer to Problem 12P

There are $2$ angles that are identical for ingle interference minima and double slit interference maxima.

Explanation of Solution

The calculation in part (a) and part (c) shows that the angles $\pm 45.7°$ are identical for both in single slit interference minima and double slit interference maxima.

So there are total $2$ angles that represent both single slit interference minima and double slit interference maxima.

Conclusion:

Therefore, there are $2$ angles that are identical for ingle interference minima and double slit interference maxima.

To determine

The number of bright fringes visible on the screen.

Answer to Problem 12P

There are $9$ bright fringes visible on the screen.

Explanation of Solution

The calculation in part (a) and part (c) shows that the angles $\pm 45.7°$ are identical for both in single slit interference minima and double slit interference maxima.

So, for the position at $\pm 45.7°$ the maxima of the double slit interference and the minima of the single slit interference minima coincide.

Thus there are $11$ maxima for the double slit interference but due to overlap of the single slit interference minima at two angles $\pm 45.7°$, the maxima are shadowed by the minima.

So, there are $11-2=9$ bright fringes visible on the screen.

Conclusion:

Therefore, there are $9$ bright fringes visible on the screen.

To determine

The intensity of the last fringe on the screen in terms of maximum intensity.

Answer to Problem 12P

The intensity of the last fringe is $0.0324I_{\max }$.

Explanation of Solution

The formula to calculate the intensity at any angle is,

$I=I_{\max }{\left[\frac{\sin \left(\frac{\pi a\sin \theta }{\lambda }\right)}{\frac{\pi a\sin \theta }{\lambda }}\right]}^2$

Here,

$I$ is the intensity at any angle.

$I_{\max }$ is the maximum intensity at the central bright.

$\theta$ is the direction of the fringe.

$a$ is the slit width.

$\lambda$ is the wavelength.

The last fringe occurs for the fifth order so the value of $\theta$ is $63.5°$.

Substitute $0.700\mu m$ for $a$ and $501.5\text{nm}$ for $\lambda$ and $63.5°$ for $\theta$ in the above equation.

$\begin{array}{c}I=I_{\max }{\left[\frac{\sin \left(\frac{\pi \left(0.700\mu \text{m}\right)\sin \left(63.5°\right)}{501.5\text{nm}}\right)}{\frac{\pi \left(0.700\mu \text{m}\right)\sin \left(63.5°\right)}{501.5\text{nm}}}\right]}^2\\ ={\mathrm{I}}_{\max }\left[\frac{\sin \left(\frac{\pi \left(0.700\mu \text{m}\times \left(\frac{1\text{m}}{{10}^6\mu \text{m}}\right)\right)\sin \left(63.5°\right)}{501.5\text{nm}\times \left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}\right)}{\frac{\pi \left(0.700\mu \text{m}\times \left(\frac{1\text{m}}{{10}^6\mu \text{m}}\right)\right)\sin \left(63.5°\right)}{501.5\text{nm}\times \left(\frac{1\text{m}}{{10}^9\text{nm}}\right)}}\right]\\ =0.0324I_{\max }\end{array}$

Conclusion:

Therefore, the intensity at the last fringe is $0.0324I_{\max }$.