$A$ single slit diffraction pattern is formed with light of wavelength $6195 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength $\lambda_0$. The value of $\lambda_0$ is (in $Å$)

$A$ plane wavefront of monochromatic light is incident normally on a narrow slit. $A$ diffraction pattern is formed on a screen. What is the phase difference between the waves originating from the top and bottom edges of the slit at the position of the first minimum?

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 $D$ is equal to:

$A$ parallel monochromatic beam of light is incident normally on a narrow slit. $A$ diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first maximum of the diffraction pattern,the phase difference between the rays coming from the edges of the slit is

$A$ screen is placed at a distance $50\,cm$ from a single slit,which is illuminated with light of wavelength $690\,nm$. If the distance between the first and third minima is $3.00\,mm$,what is the width of the slit (in $,mm$)?

$A$ beam of light of $\lambda = 600 \, nm$ from a distant source falls on a single slit $1 \, mm$ wide and the resulting diffraction pattern is observed on a screen $2 \, m$ away. The distance between the first dark fringes on either side of the central bright fringe is: