Chapter 36, Problem 18P

Monochromatic coherent light of amplitude E₀ and angular frequency ω passes through three parallel slits, each separated by a distance d from its neighbor. (a) Show that the time-averaged intensity as a function of the angle θ is

$I(\theta )=I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$

(b) Explain how this expression describes both the primary and the secondary maxima. (c) Determine the ratio of the intensities of the primary and secondary maxima. Hint: See Problem 16.

To determine

To show: The time averaged intensity as a function of angle $\theta$ is $I\left(\theta \right)=I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$.

Answer to Problem 18P

The time averaged intensity as a function of angle $\theta$ is $I\left(\theta \right)=I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$.

Explanation of Solution

Given info: The amplitude of monochromatic light is $E_0$ and angular frequency is $\omega$.

The amplitude of the monochromatic light is,

$\begin{array}{l}{E}_1=E_0\sin \omega t\\ E_2=E_0\left(\sin \omega t+\varphi \right)\\ E_3=E_0\left(\sin \omega t+2\varphi \right)\end{array}$

Here,

$\omega$ is the angular frequency.

$\varphi$ is the phase difference whose value is $\frac{2\pi d\sin \left(\theta \right)}{\lambda }$.

The total amplitude of monochromatic light is,

$E=E_1+E_2+E_3$

Substitute $E_0\sin \omega t$ for $E_1$, $E_0\left(\sin \omega t+\varphi \right)$ for $E_2$, $E_0\left(\sin \omega t+2\varphi \right)$ for $E_3$ in the above formula as,

$E=E_0\sin \omega t+E_0\left(\sin \omega t+\varphi \right)+E_0\left(\sin \omega t+2\varphi \right)$

Apply the trigonometric identity to the above expression as,

$E=2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)$+

Apply the trigonometric identity to the above expression as,

$E=2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)+E_0\sin \left(\omega t+\varphi \right)$

Add the above result to $E_2$ to get the resultant field as,

$E=2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)+E_0\sin \left(\omega t+\varphi \right)$

As it is known that intensity of monochromatic light is directly proportional to the the square of the electric field that is,

$I\propto E^2$

Here,

$I$ is the intensity.

$E$ is the electric field.

The resultant field is obtained by square the above expression as,

$\begin{array}{c}{E}^2={\left[\left[2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)\right]+\left[E_0\sin \left(\omega t+\varphi \right)\right]\right]}^2\overleftrightarrow{AB}\\ =4E^2{}_0{\cos }^2\left(\frac{\varphi }{2}\right){\sin }^2\left(\frac{2\omega t+\varphi }{2}\right)+E_0{}^2{\sin }^2\left(\omega t+\varphi \right)+4E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)E_0\sin \left(\omega t+\varphi \right)\\ =E^2{}_0{\left(1+2\cos \varphi \right)}^2\end{array}$

Substitute $\frac{2\pi d\sin \left(\theta \right)}{\lambda }$ for $\varphi$ in the above expression as,

$I\left(\theta \right)=I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$

Conclusion:

Thus, the time averaged intensity as a function of angle $\theta$ is $I\left(\theta \right)=I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$.

To determine

The way in which the expression describes both the primary and secondary maxima.

Answer to Problem 18P

The expression describes both the primary and secondary maxima.

Explanation of Solution

Given info: The amplitude of monochromatic light is $E_0$ and angular frequency is $\omega$.

From the above expression obtained in part (a), the minimum interference is obtained when $\cos \varphi$ is $\frac{1}{2}$ and maximum interference is obtained when $\cos \varphi$ is 1.

Conclusion:

Thus, the expression describes both the primary and secondary maxima.

To determine

The ratio of the intensities of the primary and secondary maxima.

Answer to Problem 18P

The ratio of the intensities of the primary and secondary maxima is $\frac{1}{9}$

Explanation of Solution

Given info: The amplitude of monochromatic light is $E_0$ and angular frequency is $\omega$.

Consider the amplitude of the monochromatic light is,

$\begin{array}{l}{E}_1=E_0\sin \omega t\\ E_2=E_0\left(\sin \omega t+\varphi \right)\\ E_3=E_0\left(\sin \omega t+2\varphi \right)\end{array}$

The total amplitude of monochromatic light is,

$E=E_1+E_2+E_3$

Substitute $E_0\sin \omega t$ for $E_1$, $E_0\left(\sin \omega t+\varphi \right)$ for $E_2$, $E_0\left(\sin \omega t+2\varphi \right)$ for $E_3$ in the above formula as,

$E=E_0\sin \omega t+E_0\left(\sin \omega t+\varphi \right)+E_0\left(\sin \omega t+2\varphi \right)$

Apply the trigonometric identity to the above expression as,

$E=2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)$+

Apply the trigonometric identity to the above expression as,

$E=2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)+E_0\sin \left(\omega t+\varphi \right)$

Add the above result to $E_2$ to get the resultant field as,

$E=2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)+E_0\sin \left(\omega t+\varphi \right)$

As it is known that intensity of monochromatic light is directly proportional to the  square of the electric field that is,

$I\propto E^2$

Here,

$I$ is the intensity.

$E$ is the electric field.

The resultant field is obtained by square the above expression as,

$\begin{array}{c}{E}^2={\left[\left[2E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)\right]+\left[E_0\sin \left(\omega t+\varphi \right)\right]\right]}^2\perp \\ =4E^2{}_0{\cos }^2\left(\frac{\varphi }{2}\right){\sin }^2\left(\frac{2\omega t+\varphi }{2}\right)+E_0{}^2{\sin }^2\left(\omega t+\varphi \right)+4E_0\cos \left(\frac{\varphi }{2}\right)\sin \left(\frac{2\omega t+\varphi }{2}\right)E_0\sin \left(\omega t+\varphi \right)\\ =E^2{}_0{\left(1+2\cos \varphi \right)}^2\end{array}$

Substitute $\frac{2\pi d\sin \left(\theta \right)}{\lambda }$ for $\varphi$ in the above expression as,

$I\left(\theta \right)=I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$

The expression for the intensity for primary maxima is,

$I\left(\theta \right)=I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$

The expression for the intensity for secondary maxima is,

$I\left(\theta \right)=9I_{\max }{\left[1+2\cos \left(\frac{2\pi d\sin \theta }{\lambda }\right)\right]}^2$

Take the ratio of the above two expression as,

$\frac{I{\left(\theta \right)}_{pri}}{I{\left(\theta \right)}_{\sec }}=\frac{1}{9}$

Conclusion:

Therefore, the ratio of the intensities of the primary and secondary maxima is $\frac{1}{9}$.