Fraunhofer Diffraction Fundamentals

Fraunhofer diffraction describes the phenomenon where light waves spread out after passing through a narrow opening or around an obstacle. In this specific problem, we are dealing with single-slit diffraction, where monochromatic light passes through a single narrow slit and produces a pattern of bright and dark fringes on a screen. The pattern consists of a bright central maximum flanked by alternating dark and bright fringes of decreasing intensity.

The position of these fringes depends on several factors: the wavelength of the light, the width of the slit, and the distance from the slit to the screen.

Dark Fringe Condition in Single-Slit Diffraction

For a single slit, the condition for destructive interference (dark fringes) is given by the formula:

$a \sin \theta = m \lambda$

• Here, a represents the width of the single slit.
• $\theta$ is the angle of the dark fringe from the center of the diffraction pattern.
• m is the order of the dark fringe (m = 1 for the first dark fringe, m = 2 for the second, and so on).
• $\lambda$ (lambda) is the wavelength of the light used.

In situations where the screen is far away from the slit and the angles are small (which is typical for Fraunhofer diffraction problems), we can use the small angle approximation:

$\sin \theta \approx \theta \approx \frac{y}{D}$

• y is the distance of the dark fringe from the center of the central bright fringe on the screen.
• D is the distance from the single slit to the screen.

Substituting the small angle approximation into the dark fringe condition, we get:

$a \left( \frac{y}{D} \right) = m \lambda$

Rearranging this formula to solve for the slit width (a):

$a = \frac{m \lambda D}{y}$

Slit Width Calculation

Let's identify the given values from the question:

• Wavelength of light, $\lambda = 580 \text{ nm}$
• Distance from the slit to the screen, $D = 1 \text{ m}$
• Distance of the first dark fringe from the central bright fringe, $y = 5.0 \text{ mm}$
• Order of the dark fringe, $m = 1$ (since it's the first dark fringe)

First, it's important to convert all units to a consistent system, preferably meters:

• $\lambda = 580 \text{ nm} = 580 \times 10^{-9} \text{ m}$
• $y = 5.0 \text{ mm} = 5.0 \times 10^{-3} \text{ m}$

Now, substitute these values into the formula for slit width:

$a = \frac{m \lambda D}{y}$

$a = \frac{(1) \times (580 \times 10^{-9} \text{ m}) \times (1 \text{ m})}{5.0 \times 10^{-3} \text{ m}}$

Perform the multiplication in the numerator:

$a = \frac{580 \times 10^{-9} \text{ m}^2}{5.0 \times 10^{-3} \text{ m}}$

Now, divide the values and handle the powers of 10:

$a = \left( \frac{580}{5.0} \right) \times 10^{(-9) - (-3)} \text{ m}$

$a = 116 \times 10^{-6} \text{ m}$

To express this value in millimeters (mm), recall that $1 \text{ mm} = 10^{-3} \text{ m}$:

$a = 116 \times 10^{-3} \times 10^{-3} \text{ m}$

So, $a = 116 \times 10^{-3} \text{ mm}$

$a = 0.116 \text{ mm}$

Rounding this value to two significant figures, or the precision typically found in the options, gives:

$a \approx 0.12 \text{ mm}$

Therefore, the slit width is approximately 0.12 mm.