QJEEM24Phdbbd: JEE Main 2024 Physics - Class 12 JEEM24Phdbbd Solution: A microwave of wavelength falls normally on a slit of width .…

To determine the angular spread of the central maximum in a single-slit diffraction pattern, we can use the formula for the angular position of the first minimum (also understood as the boundary of the central maximum) on either side of the center. For a single-slit diffraction pattern, the angle $\theta$ to the first minimum is given by the condition:

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

where :

• $a$ is the width of the slit,


• $\lambda$ is the wavelength of the light (or microwaves in this case),


• $m$ is the order number of the minimum with $m = \pm1, \pm2, \pm3, ...$ (for the first minimum, $m = \pm1$ )

However, we are only interested in the angle to the first minimum, so we will only consider $m = \pm1$ . Since the slit width $a = 4.0 \mathrm{cm}$ and the wavelength $\lambda = 2.0 \mathrm{cm}$ , we substitute these values into the equation to find $\theta$ :

$4.0 \mathrm{cm} \times \sin(\theta) = 1 \times 2.0 \mathrm{cm}$

$\Rightarrow$ $4.0 \sin(\theta) = 2.0$

$\Rightarrow$ $\sin(\theta) = \frac{2.0}{4.0}$

$\Rightarrow$ $\sin(\theta) = 0.5$

The angle whose sine is 0.5 is $30^{\circ}$ .

This angle of $30^{\circ}$ is the angle from the center to the first minimum on one side. The angular spread of the central maximum would cover the range from the first minimum on one side to the first minimum on the other side, totalling twice this angle:

$\text{Angular spread} = 2 \times \theta = 2 \times 30^{\circ} = 60^{\circ}$

Therefore, the angular spread of the central maxima of the diffraction pattern is $60^{\circ}$ , which corresponds to Option A.