QJEEM24Ph18ea: JEE Main 2024 Physics - Class 12 JEEM24Ph18ea Solution: A monochromatic light of wavelength is incident on the single…

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.