In a double slit experiment,the two slits are $1\, mm$ apart and the screen is placed $1\, m$ away. $A$ monochromatic light of wavelength $500\, nm$ is used. What will be the width of each slit for obtaining ten maxima of the double slit pattern within the central maxima of the single slit pattern? .......$mm$

$A$ 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:

In the diffraction pattern, the first maximum is at $30^{\circ}$, when a monochromatic light of wavelength $\lambda$ is incident on a slit of width $a$. For the same wavelength, if the slit width is changed, so that the first maximum is at $45^{\circ}$, by how much is the slit width changed?

$A$ slit of size $0.15 \,cm$ is placed at $2.1 \,m$ from a screen. On illuminating it by a light of wavelength $5 \times 10^{-5} \,cm$,the width of the central maxima will be:

$A$ 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)?

$A$ parallel beam of light of wavelength $\lambda$ 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 second minimum of the diffraction pattern,the phase difference between the rays coming from the two edges of the slit is: