$A$ plane wavefront $(\lambda = 6 \times 10^{-7} \, m)$ falls on a slit $0.4 \, mm$ wide. $A$ convex lens of focal length $0.8 \, m$ placed behind the slit focuses the light on a screen. What is the linear diameter of the second maximum 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:

$A$ single slit diffraction experiment is performed to determine the slit width using the equation,$\frac{b d}{D} = m \lambda$,where $b$ is the slit width,$D$ is the distance between the slit and the screen,$d$ is the distance between the $m^{\text{th}}$ diffraction maximum and the central maximum,and $\lambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of $1 \ cm$ and $1 \ mm$,respectively. The values of $\lambda$ and $m$ are known precisely to be $600 \ nm$ and $3$,respectively. The absolute error (in $\mu m$) in the value of $b$ estimated using the diffraction maximum that occurs for $m=3$ with $d=5 \ mm$ and $D=1 \ m$ is $.....$

$A$ screen is placed at $50 \ cm$ from a single slit,which is illuminated with light of wavelength $600 \ nm$. If the separation between the $1^{st}$ and $3^{rd}$ minima in the diffraction pattern is $3 \ mm$,then the slit width is: (in $mm$)

In a single slit diffraction experiment,the width of the slit is made double its original width. Then the central maximum of the diffraction pattern will become

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: