A single slit of width $$b$$ is illuminated by a coherent monochromatic light of wavelength $$\lambda$$. If the second and fourth minima in the diffraction pattern at a distance 1 cm from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)

For diffraction of light by a single slit, the positions of the minima are obtained from the condition

$$b \sin\theta_m = m\lambda,$$

where $$b$$ is the slit width, $$\lambda$$ is the wavelength, $$\theta_m$$ is the angle that the $$m^{\text{th}}$$ minimum makes with the central axis, and $$m = 1,2,3,\ldots$$ labels the order of the minimum.

The screen is placed at a distance $$D$$ from the slit. When the angles are small, we use the small-angle approximations

$$\tan\theta_m \approx \sin\theta_m \approx \theta_m \quad\text{(in radians)},$$

so the transverse distance $$y_m$$ of the $$m^{\text{th}}$$ minimum from the central maximum on the screen is

$$y_m = D\tan\theta_m \;\approx\; D\sin\theta_m.$$

Substituting $$\sin\theta_m = m\lambda/b$$ from the diffraction condition, we get an explicit formula for the position of the minima on the screen:

$$y_m \approx D\left(\frac{m\lambda}{b}\right) = \frac{D\lambda}{b}\,m.$$

This shows that, under the small-angle approximation, the distances of successive minima are directly proportional to the order $$m$$.

We are told that

$$y_2 = 3\ \text{cm}, \qquad y_4 = 6\ \text{cm}.$$

To confirm consistency, notice that

$$\frac{y_4}{y_2} = \frac{6\ \text{cm}}{3\ \text{cm}} = 2 = \frac{4}{2},$$

which matches the proportionality predicted by the formula, so the data are self-consistent.

From the expression $$y_m = (D\lambda/b)\,m,$$ solve for the constant factor $$D\lambda/b$$ using the second-order minimum (any order would work):

$$\frac{D\lambda}{b} = \frac{y_2}{2} = \frac{3\ \text{cm}}{2} = 1.5\ \text{cm}.$$

Now, the first-order minima (which bound the central maximum) correspond to $$m = 1$$. Their distances from the central axis are therefore

$$y_1 = \frac{D\lambda}{b}\, (1) = 1.5\ \text{cm}.$$

The central bright fringe extends from the first minimum on one side to the first minimum on the other side, so its full width is twice $$y_1$$:

$$\text{Width of central maximum} = 2y_1 = 2 \times 1.5\ \text{cm} = 3.0\ \text{cm}.$$

Hence, the correct answer is Option D.