Question

You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80.0 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance between them to be 17.9millimeters. (Figure 1)

What is the width a of the slit? Give your answer in micrometers, to three sig figs.

Answer 1

Answer 2

Concepts and reason

The concept of single slit diffraction is required to solve the problem.

Initially use the trigonometry ratio of tangent. Use the small angle approximation as the screen is very far from the slit. Finally, determine the slit width by using the single slit diffraction condition for minimum.

Fundamentals

The condition for minimum in the single slit diffraction is,

$a\sin \theta = m\lambda$

Here, a is the slit width, $\lambda$ is the wavelength of light, m is the order of fringe, and $\theta$ is the angle of diffraction.

The trigonometry ratio for tangent is,

$\tan \theta = \frac{y}{L}$

Here, y is the distance from the center to the dark fringe and L is the distance between the screen and the slit.

Consider the screen is very far from the slit.

As the screen is very far from the slit, diffraction angle will be very small. Therefore, for small diffraction angle, the sine of diffraction angle is given as,

$\sin \theta = \tan \theta$

Substitute $\frac{y}{L}$ for $\tan \theta$ in equation$\sin \theta = \tan \theta$.

$\sin \theta = \frac{y}{L}$

From the figure, the given two minima are equal distance from the center maximum. Distance between the two minima (D) is,

$D = 2y$

Rearrange the equation for y.

$y = \frac{D}{2}$

Determine the width of the slit.

The condition for single slit diffraction minimum is,

$a\sin \theta = m\lambda$

Substitute $\frac{y}{L}$ for $\sin \theta$ in equation $a\sin \theta = m\lambda$ and rearrange the equation for d.

$\begin{array}{c}\\a\left( {\frac{y}{L}} \right) = m\lambda \\\\a = \frac{{m\lambda L}}{y}\\\end{array}$

Substitute $\frac{D}{2}$ for y in the equation$a = \frac{{m\lambda L}}{y}$.

$a = \frac{{2m\lambda L}}{D}$

Substitute 3 for m, 633 nm for$\lambda$, 80.0 cm for L, and 17.9 mm for D in the above equation $a = \frac{{2m\lambda L}}{D}$ and determine the width of the slit.

$\begin{array}{c}\\a = \frac{{2\left( 3 \right)\left( {633{\rm{ nm}}\left( {\frac{{{{10}^{ - 9}}{\rm{ m}}}}{{1{\rm{ nm}}}}} \right)} \right)\left( {80.0{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1\,{\rm{cm}}}}} \right)} \right)}}{{\left( {17.9{\rm{ mm}}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)} \right)}}\\\\ = 1.70 \times {10^{ - 4}}{\rm{ m}}\left( {\frac{{{{10}^6}{\rm{ }}\mu {\rm{m}}}}{{1{\rm{ m}}}}} \right)\\\\ = 170{\rm{ }}\mu {\rm{m}}\\\end{array}$

Ans:

The width of the slit is$170{\rm{ }}\mu {\rm{m}}$.