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

<p>To find the linear width of the central maximum in a single-slit diffraction pattern, we can use the formula that relates the position of the first minima on either side of the central maximum. The angle, $ \theta $, at which the first minimum occurs is given by:</p> <p>$a \sin(\theta) = m\lambda$</p> <p>Where:</p> <ul> <li>$ a $ is the width of the slit</li><br> <li>$ \lambda $ is the wavelength of monochromatic light</li><br> <li>$ m $ is the order number of the minimum (for the first minimum, $ m = \pm 1 $)</li> </ul> <p>In our case, we want to find the position of the first minima to determine the width of the central maximum on the screen. So we will use $ m = 1 $ and $ m = -1 $ which correspond to the first minima on either side of the central peak.</p> <p>Given the width of the slit $ a = 0.01 $ mm, which we need to convert to meters for consistency with the wavelength:</p> <p>$a = 0.01 \times 10^{-3} \text{m}$</p> <p>And the wavelength $ \lambda = 6000 \mathring{A} $, also converting to meters:</p> <p>$\lambda = 6000 \times 10^{-10} \text{m}$</p> <p>We're using a convex lens of focal length $ f = 20 $ cm to project the pattern onto a screen. We'll need to convert the focal length to meters as well:</p> <p>$f = 20 \times 10^{-2} \text{m}$</p> <p>We can calculate the angle $ \theta $ needed for the first minima using the approximation of small angles, where $ \sin(\theta) \approx \theta $:</p> <p>$a \theta = m\lambda$</p> <p>Now, solve for $ \theta $ for the first minimum (m = 1):</p> <p>$\theta = \frac{\lambda}{a}$</p> <p>Plugging in the values:</p> <p>$\theta = \frac{6000 \times 10^{-10}}{0.01 \times 10^{-3}}$</p> <p>$\theta = \frac{6000 \times 10^{-10}}{10^{-5}}$</p> <p>$\theta = 6000 \times 10^{-5}$</p> <p>Now, linear width of the central maximum on the screen (from -m to m, or -1 to +1) will be twice the distance from the center to the first minimum, which can be found using the focal length of the lens $ f $ and the angle $ \theta $:</p> <p>$y = 2f\theta$</p> <p>Plugging the focal length and calculated theta:</p> <p>$y = 2 \times 20 \times 10^{-2} \times 6000 \times 10^{-5}$</p> <p>$y = 40 \times 10^{-2} \times 6000 \times 10^{-5}$</p> <p>$y = 240 \times 10^{-3} \text{m}$</p> <p>$y = 24 \times 10^{-2} \text{m}$</p> <p>$y = 24 \text{mm}$</p> <p>This means the linear width of the central maximum is 24 mm. Hence, the correct answer is Option B.</p>