JEE MAIN - Physics (2024 - 1st February Morning Shift - No. 6)

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:

$$ a \sin(\theta) = m\lambda $$

Where:

• $ a $ is the width of the slit


• $ \lambda $ is the wavelength of monochromatic light


• $ m $ is the order number of the minimum (for the first minimum, $ m = \pm 1 $)

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.

Given the width of the slit $ a = 0.01 $ mm, which we need to convert to meters for consistency with the wavelength:

$$ a = 0.01 \times 10^{-3} \text{m} $$

And the wavelength $ \lambda = 6000 \mathring{A} $, also converting to meters:

$$ \lambda = 6000 \times 10^{-10} \text{m} $$

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:

$$ f = 20 \times 10^{-2} \text{m} $$

We can calculate the angle $ \theta $ needed for the first minima using the approximation of small angles, where $ \sin(\theta) \approx \theta $:

$$ a \theta = m\lambda $$

Now, solve for $ \theta $ for the first minimum (m = 1):

$$ \theta = \frac{\lambda}{a} $$

Plugging in the values:

$$ \theta = \frac{6000 \times 10^{-10}}{0.01 \times 10^{-3}} $$

$$ \theta = \frac{6000 \times 10^{-10}}{10^{-5}} $$

$$ \theta = 6000 \times 10^{-5} $$

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 $:

$$ y = 2f\theta $$

Plugging the focal length and calculated theta:

$$ y = 2 \times 20 \times 10^{-2} \times 6000 \times 10^{-5} $$

$$ y = 40 \times 10^{-2} \times 6000 \times 10^{-5} $$

$$ y = 240 \times 10^{-3} \text{m} $$

$$ y = 24 \times 10^{-2} \text{m} $$

$$ y = 24 \text{mm} $$

This means the linear width of the central maximum is 24 mm. Hence, the correct answer is Option B.