Question

After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.0⁰with the original direction of the beam, as veiwed on a screen farfrom the slits.

(a) What is the ratio of the distance between the slits to thewavelength of the light illuminating the slits?

(b) What is the smallest angle, relative to original directionof the laser beam, at which the intensity of the light is 1/10 themaximum intensity on the screen?

Answer 1

Concepts and reason

The concept used to solve this problem is the destructive interference of waves.

First, use the condition for destructive interference to determine the ratio of the distance between the slits to the wavelength of the light.

Then, use the expression for the intensity of light to determine the smallest angle at which the intensity of light is $1/{10^{{\rm{th}}}}$, which is the maximum intensity on the screen.

Fundamentals

Interference is the phenomenon in which two or more waves superpose to form a resultant wave of greater, lesser or the same amplitude.

Destructive interference occurs when the amplitude of the interfering waves subtract so that the resultant wave will have a lesser amplitude than the interfering waves.

The condition for destructive interference is given below:

$\delta = \left( {m + \frac{1}{2}} \right)\lambda$

Here, $\delta$ is the path difference between the interfering waves, $m$ is the order of the interference pattern and $\lambda$ is the wavelength.

The condition for destructive interference given by Young’s double slit experiment is given below:

$d\sin \theta = \left( {m + \frac{1}{2}} \right)\lambda$

Here, $d$ is the distance between the slits, $\theta$ is the angle the rays make relative to a perpendicular line joining the slits to the screen.

The expression for the intensity of the light is given below:

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

Here, $I$ is the intensity of the light, ${I_0}$ is the maximum intensity on the screen, and $\phi$ is the phase shift.

The expression for phase shift in terms of path difference is given below:

$\phi = \frac{{2\pi }}{\lambda }\delta$

(a)

The condition for destructive interference given by Young’s double slit experiment is given below:

$d\sin \theta = \left( {m + \frac{1}{2}} \right)\lambda$

The ratio of the distance between the slits to the wavelength of the light is given below:

$\frac{d}{\lambda } = \frac{{\left( {m + \frac{1}{2}} \right)}}{{\sin \theta }}$

Substitute $0$ for $m$ and $15.0^\circ$ for $\theta$.

$\begin{array}{c}\\\frac{d}{\lambda } = \frac{{\left( {0 + \frac{1}{2}} \right)}}{{\sin 15.0^\circ }}\\\\ = 1.93\\\end{array}$

(b)

The expression for the intensity of the light is given below:

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

The expression for phase shift in terms of path difference is given below:

$\phi = \frac{{2\pi }}{\lambda }\delta$

Substitute $\left( {2\pi /\lambda } \right)\delta$ for $\phi$ in the above expression for the intensity of light:

$\begin{array}{c}\\I = {I_0}{\cos ^2}\left( {\frac{{\left( {\frac{{2\pi }}{\lambda }\delta } \right)}}{2}} \right)\\\\ = {I_0}{\cos ^2}\left( {\frac{{\pi \delta }}{\lambda }} \right)\\\end{array}$

Substitute $d\sin \theta$ for $\delta$ in the above expression:

$I = {I_0}{\cos ^2}\left( {\frac{{\pi d\sin \theta }}{\lambda }} \right)$

Rearrange the above expression to obtain the formula for the angle.

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

Thus, the expression for the angle is given below:

$\theta = {\sin ^{ - 1}}\left( {\left( {\frac{\lambda }{{\pi d}}} \right)\left( {{{\cos }^{ - 1}}\sqrt {\frac{I}{{{I_0}}}} } \right)} \right)$

Substitute $\left( {1/1.93} \right)$ for $\left( {\lambda /d} \right)$, $3.14$ for $\pi$, and $\left( {1/10} \right){I_0}$ for $I$:

$\begin{array}{c}\\\theta = {\sin ^{ - 1}}\left( {\left( {\left( {\frac{1}{{1.93}}} \right)\left( {\frac{1}{{3.14}}} \right)} \right)\left( {{{\cos }^{ - 1}}\sqrt {\frac{{\frac{1}{{10}}{I_0}}}{{{I_0}}}} } \right)} \right)\\\\ = {\sin ^{ - 1}}\left( {\left( {0.165} \right)\left( {71.565^\circ } \right)} \right)\\\\ = {\sin ^{ - 1}}\left( {\left( {0.165} \right)\left( {71.565^\circ \left( {\frac{{\left( {\frac{\pi }{{180}}{\rm{rad}}} \right)}}{{1^\circ }}} \right)} \right)} \right)\\\\ = {\sin ^{ - 1}}\left( {0.206} \right)\\\\ = 11.89^\circ \\\end{array}$

Ans: Part a

The ratio of the distance between the slits to the wavelength of the light is ${\bf{1}}{\bf{.93}}$.

Part b

Thus, the smallest angle is ${\bf{11}}{\bf{.89^\circ }}$.