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Step-by-Step Detailed Solution
Step 1: Identify the Diffraction Condition
In a single-slit diffraction experiment, for the first minima on either side of the central bright fringe, the condition is:
$ a \sin(\theta) = m \lambda $,
where:
$ a $ is the slit width
$ \lambda $ is the wavelength of light
$ m = \pm 1, \pm 2, \ldots$ (for the first minima, $m = \pm 1$)
The central maximum is the region between these first minima, spanning from $-\theta$ to $+\theta$.
Step 2: Convert the Given Values into SI Units
Slit width:
$ a = 0.01 \,\text{mm} = 0.01 \times 10^{-3}~\text{m} = 1.0 \times 10^{-5}~\text{m} $
Wavelength:
$ \lambda = 6000 \,\mathring{A} = 6000 \times 10^{-10}~\text{m} = 6.0 \times 10^{-7}~\text{m} $
Focal length of the convex lens:
$ f = 20 \,\text{cm} = 20 \times 10^{-2}~\text{m} = 0.20~\text{m} $
Step 3: Calculate the Diffraction Angle for the First Minimum
For small angles (in radians), $ \sin(\theta) \approx \theta $. Thus,
$$
a \sin(\theta) = \lambda \quad \Longrightarrow \quad \theta \approx \frac{\lambda}{a}.
$$
Substituting the values:
$$
\theta = \frac{6.0 \times 10^{-7}}{1.0 \times 10^{-5}} = 6.0 \times 10^{-2} = 0.06~\text{radians}.
$$
Step 4: Determine the Linear Width of the Central Maximum
The total angular width of the central maximum is $ 2\theta $. When the diffraction pattern is formed at the focal plane of a lens of focal length $ f $, the linear width $ y $ is:
$$
y = 2 f \theta.
$$
Substituting $ f = 0.20~\text{m} $ and $ \theta = 0.06 $:
$$
y = 2 \times 0.20 \times 0.06 = 0.024~\text{m} = 24~\text{mm}.
$$
Step 5: Final Answer
The linear width of the central diffraction maximum is
$24 \mathrm{~mm}$.
