A monochromatic light of wavelength $$6000\ \mathring{A}$$ is incident on the single slit of width $$0.01$$ mm. If the diffraction pattern is formed at the focus of the convex lens of focal length $$20$$ cm, the linear width of the central maximum is :

Find the linear width of the central maximum in single-slit diffraction.

In single-slit diffraction the first minima occur at angles where $$a\sin\theta = \pm\lambda$$. For small angles ($$\sin\theta \approx \theta$$) the distance from the center to the first minimum on the screen is given by $$y = \frac{\lambda f}{a}$$, where $$f$$ is the focal length of the lens, so the total width of the central maximum is $$2y = \frac{2\lambda f}{a}$$.

The wavelength is $$\lambda = 6000\text{ A} = 6000 \times 10^{-10}\text{ m} = 6 \times 10^{-7}\text{ m}$$, the slit width is $$a = 0.01\text{ mm} = 0.01 \times 10^{-3}\text{ m} = 1 \times 10^{-5}\text{ m}$$, and the focal length is $$f = 20\text{ cm} = 0.2\text{ m}$$.

Substituting these values into the expression for the width gives $$2y = \frac{2 \times 6 \times 10^{-7} \times 0.2}{1 \times 10^{-5}} = \frac{2.4 \times 10^{-7}}{10^{-5}} = 2.4 \times 10^{-2}\text{ m} = 24\text{ mm}$$.

The correct answer is Option B: 24 mm.