In Fraunhofer diffraction pattern, slit width is 0.3 mm and screen is at 1.5 m away from the lens. If wavelength of light used is $4500 Å$, then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.]

The angular position $\theta_1$ of the first diffraction minimum is given by $a\sin\theta_1 = \lambda$, where $a = 0.3 \times 10^{-3}$ m and $\lambda = 4.5 \times 10^{-7}$ m.
Using the small-angle approximation $\sin\theta_1 \approx \theta_1$, this simplifies to $\theta_1 = \lambda/a$.

The linear distance from the central maximum to the first minimum is then $y_1 = D\theta_1 = D\lambda/a$, with $D = 1.5$ m.
The separation between the first minima on either side is $2y_1 = 2D\lambda/a$.

Substituting the values: $2 \times 1.5 \times \frac{4.5 \times 10^{-7}}{0.3 \times 10^{-3}} = 45 \times 10^{-4} = 4.5 \times 10^{-3}$ m.

The distance is 4.5 mm, corresponding to option D.