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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Slit width, $a = 0.1 \text{ mm} = 1 \times 10^{-4} \text{ m}$
• Wavelength of light, $\lambda = 6000 \, \text{Å} = 6 \times 10^{-7} \text{ m}$
• Distance from slit to screen, $D = 0.5 \text{ m}$
• Order of dark band (minima) we want: $n = 3$
Step 2: Write the Condition for Dark Bands in Single-Slit Diffraction
The formula for the position of the $n$-th dark band (minima) in single-slit diffraction is:
$$
a \,\sin \theta = n\,\lambda
$$
Considering small angles (valid for typical diffraction experiments), $\sin \theta \approx \frac{x}{D}$,
where $x$ is the distance of the dark fringe from the central maximum on the screen.
Hence, we have:
$$
a \cdot \frac{x}{D} = n \,\lambda
$$
Step 3: Substitute the Values to Find $x$
From
$$
\frac{a \, x}{D} = n \,\lambda \quad \Longrightarrow \quad
x = \frac{n \,\lambda \, D}{a}
$$
Substituting $n = 3$, $\lambda = 6 \times 10^{-7} \text{ m}$, $D = 0.5 \text{ m}$, and $a = 1 \times 10^{-4} \text{ m}$:
$$
x = \frac{3 \times \bigl(6 \times 10^{-7}\bigr) \times 0.5}{1 \times 10^{-4}}
$$
$$
x = \frac{3 \times 6 \times 10^{-7} \times 0.5}{10^{-4}}
= \frac{9 \times 10^{-7}}{10^{-4}} = 9 \times 10^{-3} \text{ m} = 9 \text{ mm}
$$
Step 4: State the Final Answer
Therefore, the distance of the third dark band from the central bright band is
9 mm.
