JEE Main 2024 — Wave Optics Question with Solution

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.