In a single slit diffraction experiment, for slit width '$\alpha$' the width of the central maxima is '$\beta$'. If we double the slit width then the corresponding width of the central maxima will be:

The angular width of the central maximum in a single slit diffraction experiment is given by the formula $\theta = \dfrac{2\lambda}{a}$, where $a$ is the slit width and $\lambda$ is the wavelength of light.

The linear width of the central maximum on a screen at a distance $D$ from the slit is given by $\beta = D \times \theta = \dfrac{2\lambda D}{a}$.

From this expression, it is evident that the width of the central maximum $\beta$ is inversely proportional to the slit width $a$, i.e., $\beta \propto \dfrac{1}{a}$.

Let the initial slit width be $a_1 = a$ and the initial width of the central maximum be $\beta_1 = \beta$.

When the slit width is doubled, the new slit width $a_2 = 2a$.

The new width of the central maximum $\beta_2$ is given by $\beta_2 = \dfrac{2\lambda D}{a_2} = \dfrac{2\lambda D}{2a} = \dfrac{1}{2} \left( \dfrac{2\lambda D}{a} \right) = \dfrac{\beta}{2}$.

Answer: $\dfrac{\beta}{2}$