Step 1: Understanding the Concept:
In single-slit Fraunhofer diffraction, the central maximum is flanked by minima on both sides.
The distance between the first minimum on either side is effectively the linear width of the central maximum.
We can use the formula for the position of diffraction minima to find the wavelength of the light.
Step 2: Key Formula or Approach:
The linear position $y_n$ of the $n^{\text{th}}$ minimum on the screen is $y_n = \frac{n \lambda D}{a}$, where $\lambda$ is wavelength, $D$ is distance to the screen, and $a$ is the slit width.
The total distance between the first minima on either side is $2y_1 = \frac{2\lambda D}{a}$.
Step 3: Detailed Explanation:
Given slit width $a = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$.
Distance to screen $D = 2 \text{ m}$.
Distance between first minima on either side $= 2y_1 = 1 \text{ cm} = 10^{-2} \text{ m}$.
Set up the equation using the formula for the width of the central maximum:
\[ 2y_1 = \frac{2\lambda D}{a} \]
Substitute the given values into the equation:
\[ 10^{-2} = \frac{2 \cdot \lambda \cdot 2}{0.2 \times 10^{-3}} \]
Simplify the equation to solve for $\lambda$:
\[ 10^{-2} = \frac{4\lambda}{0.2 \times 10^{-3}} \]
Multiply both sides by the denominator:
\[ 4\lambda = 10^{-2} \times 0.2 \times 10^{-3} \]
\[ 4\lambda = 0.2 \times 10^{-5} \]
\[ 4\lambda = 2 \times 10^{-6} \]
Divide by 4:
\[ \lambda = \frac{2 \times 10^{-6}}{4} = 0.5 \times 10^{-6} \text{ m} \]
To match the options, convert meters to Angstroms ($1 \text{ m} = 10^{10} \text{ \AA}$):
\[ \lambda = 0.5 \times 10^{-6} \times 10^{10} \text{ \AA} \]
\[ \lambda = 0.5 \times 10^4 \text{ \AA} \]
\[ \lambda = 5000 \text{ \AA} \]
Step 4: Final Answer:
The wavelength of light used is $5000 \text{ \AA}$.