In the case of Fraunhofer diffraction due to a single slit,the intensity of the central maximum is $I_0$. If the width of the slit is doubled,what will be the new intensity of the central maximum?

Light of wavelength $\lambda$ is incident on a single slit of width $a$,and the distance between the slit and the screen is $D$. In the diffraction pattern,if the slit width is equal to the width of the central maximum,then $D=$

Light of wavelength $5000 \text{ Å}$ is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of $5 \text{ mm}$ from the central maximum on a screen placed at a distance of $2 \text{ m}$ from the slit. The width of the slit is: (in $\text{ cm}$)

The width of the central maximum of a diffraction pattern on a single slit does not depend upon:

$A$ screen is placed $50\,cm$ from a single slit,which is illuminated with $6000\,\mathring{A}$ light. If the distance between the first and third minima in the diffraction pattern is $3\,mm,$ the width of the slit is ......$mm$.

In a single slit diffraction experiment, the first minimum for red light $(660 \ nm)$ coincides with the first secondary maximum of some other wavelength $\lambda$. The value of $\lambda$ is $..... \mathring{A}$.