Chapter 35, Problem 89PQ

A One of the slits in a Young’s double-slit apparatus is wider than the other, so that the amplitude of the light that reaches the central point of the screen from one slit alone is twice that from the other slit alone. Determine the resultant intensity as a function of the direction $\theta$ on the screen, the wavelength $\lambda$ of the incident light, the incident intensity I₀, and the slit separation d.

To determine

The resultant intensity as a function of the wavelength of the incident light $\lambda$, incident intensity is $I_0$ and the angle $\theta$.

Answer to Problem 89PQ

The resultant intensity as a function of the wavelength of the incident light $\lambda$, incident intensity is $I_0$ and the angle $\theta$ is $I=\frac{I_0}{9}\left(1+8{\cos }^2\left(\frac{\pi }{\lambda }d\sin \theta \right)\right)$.

Explanation of Solution

Given that the one of the slit in the Young’s double slit apparatus is wider than the other, so that the amplitude of the light on the central portion is twice alone than the other.

Write the relation of the intension anywhere on the screen.

    $I=I_0{\cos }^2\left(\frac{\phi }{2}\right)$                                                                                (I)

Where,

    $I_0=\frac{E_0{}^2}{2{\mu }_{0}c}$                                                                                       (II)

    $\phi =\frac{2\pi }{\lambda }d\sin \theta$                                                                                   (III)

Here, $\phi$ is the phase difference, $I_0$ is the maximum intensity, $\lambda$ is wavelength of light used, $d$ is the distance between the two slits, $E_0$ is the amplitude of the electric field, $c$ is the speed of light, $\mu$ is the permeability of the free space.

The amplitude of light from first slit.

    $E_{\text{A}}=E_1\sin \omega t$

The amplitude of the light from other slit.

    $E_{\text{B}}=E_2\sin \left(\omega t+\phi \right)$

Here, $\omega$ is the angular frequency.

The resultant amplitude $E_0$.

    $E_0=\sqrt{E^2{}_1+E^2{}_2+2E_1{E}_2\cos \phi }$                                                                        (IV)

Conclusion:

Substitute, $2E_1$ for $E_2$ in equation (IV) to find $E_1$

    $\begin{array}{c}{E}_0=\sqrt{E^2{}_1+{\left(2E_1\right)}^2+2E_1\left(2E_1\right)\cos \phi }\\ =\sqrt{E^2{}_1+4E_1{}^2+4E_1{}^2\cos \phi }\end{array}$

Take square both the sides.

    $E_0{}^2=E^2{}_1+4E_1{}^2+4E_1{}^2\cos \phi$                                                                        (V)

Take $\phi =0°$ for central fringe.

    $\begin{array}{c}{E}_0{}^2=E^2{}_1+4E_1{}^2+4E_1{}^2\cos 0°\\ E_0{}^2=E^2{}_1+4E_1{}^2+4E_1{}^2\\ E_0{}^2=9E^2{}_1\\ E^2{}_1=\frac{E^2{}_0}{9}\end{array}$                                                                      (VI)

Substitute, $E^2{}_1+4E_1{}^2+4E_1{}^2\cos \phi$ for $E^2{}_0$ from equation (V) in equation (II) to find $I$.

    $I=\frac{1}{2{\mu }_{0}c}\left(E^2{}_1+4E_1{}^2+4E_1{}^2\cos \phi \right)$

Substitute, $\frac{E^2{}_0}{9}$ for $E^2{}_1$ from equation (VI) in the above relation.

    $\begin{array}{c}I=\frac{1}{2{\mu }_{0}c}\left(\frac{E^2{}_0}{9}+4\times \frac{E^2{}_0}{9}+4\times \frac{E^2{}_0}{9}\cos \phi \right)\\ =\frac{E^2{}_0}{9\times 2{\mu }_{0}c}\left(1+4+4\cos \phi \right)\end{array}$

Substitute, $\frac{E^2{}_0}{2{\mu }_{0}c}$ for $I_0$ from equation (III) in the above equation.

  $\begin{array}{c}I=\frac{I_0}{9}\left(1+4+4\cos \phi \right)\\ =\frac{I_0}{9}\left(1+4+4\left(2{\cos }^2\left(\frac{\phi }{2}\right)-1\right)\right)\\ =\frac{I_0}{9}\left(1+4+8{\cos }^2\left(\frac{\phi }{2}\right)-4\right)\\ =\frac{I_0}{9}\left(1+8{\cos }^2\left(\frac{\phi }{2}\right)\right)\end{array}$

Substitute, $\frac{2\pi }{\lambda }d\sin \theta$ for $\phi$ from equation (II) in the above equation.

    $\begin{array}{c}I=\frac{I_0}{9}\left(1+8{\cos }^2\left(\frac{\frac{2\pi }{\lambda }d\sin \theta }{2}\right)\right)\\ =\frac{I_0}{9}\left(1+8{\cos }^2\left(\frac{\pi }{\lambda }d\sin \theta \right)\right)\end{array}$

Therefore, the resultant intensity as a function of the wavelength of the incident light $\lambda$ , incident intensity is $I_0$ and the angle $\theta$ is $I=\frac{I_0}{9}\left(1+8{\cos }^2\left(\frac{\pi }{\lambda }d\sin \theta \right)\right)$.