Light of wavelength $\lambda = 5000 \ \mathring{A}$ falls normally on a narrow slit. $A$ screen is placed at a distance of $1 \ m$ from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at $5 \ mm$ from the centre of the central maximum. The width of the slit is ....... $mm$.

$A$ 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 $1\,m$ from the slit are at $3\,cm$ and $6\,cm$ respectively from the central maximum,what is the width of the central maximum in $cm$ (i.e.,distance between the first minimum on either side of the central maximum)?

Through a narrow slit of width $2 \,mm$, a diffraction pattern is formed on a screen kept at a distance $2 \,m$ from the slit. The wavelength of the light used is $6330 \mathring{A}$ and it falls normal to the slit and screen. Then, the distance between the two minima on either side of the central maximum is (in $\,mm$)

In a Fraunhofer's diffraction obtained by a single slit aperture,the value of path difference for $n^{th}$ order of minima is

In a single-slit diffraction experiment,what is the width of the slit $(b)$ such that $20$ maxima of a double-slit interference pattern are obtained within the central maximum of the diffraction pattern? (Given: Slit separation for the double-slit arrangement $d = 2 \, mm$)

In a single slit diffraction pattern,if the first maxima of light of wavelength $\lambda$ coincides with the first minima of yellow light $(540 \ nm)$,then $\lambda$ is ......... $nm$.