ght be $\lambda$.

Step 1: Condition for minima in single-slit Fraunhofer diffraction

For the first minimum (on either side), the angular position $\theta$ satisfies:

$$ a \sin\theta = m\lambda, \quad \text{where } m = \pm 1 $$

Step 2: Small angle approximation

For small $\theta$, $\sin\theta \approx \tan\theta \approx \theta$.

The linear distance from the center to the first minimum on the screen,

$$ y_1 = D \tan\theta \approx D \sin\theta \approx D \theta $$

So,

$$ \sin\theta = \frac{y_1}{D} $$

Step 3: Distance between first minima on both sides

First minimum appears at $y_1$ above and $-y_1$ below the central maximum, so the total distance is

$$ Y = 2y_1 $$

Given $Y = 0.01\, \text{m}$,

$$ y_1 = \frac{Y}{2} = \frac{0.01}{2} = 0.005\,\text{m} $$

Step 4: Substitute and solve

Using $a \sin\theta = \lambda$,

$$ a \sin\theta = \lambda $$

$$ \lambda = a \sin\theta = a \frac{y_1}{D} $$

Substitute the values:

$a = 2 \times 10^{-4}~\text{m}$
$y_1 = 0.005~\text{m}$
$D = 2~\text{m}$

So,

$$ \lambda = 2 \times 10^{-4} \times \frac{0.005}{2} $$

$$ \lambda = 2 \times 10^{-4} \times 0.0025 $$

$$ \lambda = 5 \times 10^{-7}~\text{m} $$

Step 5: Convert to Angstrom

$1~\text{Å} = 10^{-10}~\text{m}$

$$ \lambda = 5 \times 10^{-7}~\text{m} = 5 \times 10^{-7}~\text{m} \times \frac{1~\text{Å}}{10^{-10}~\text{m}} = 5 \times 10^{3}~\text{Å} = 5000~\text{Å} $$

Final Answer

Option C: $5000~\mathring{A}$

Wave Optics

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