For Bragg's diffraction by a crystal to occur,the $X$-ray of wavelength $\lambda$ and interatomic distance $d$ must satisfy which condition?

The condition for obtaining secondary maxima in the diffraction pattern due to a single slit is:

$A$ narrow slit of width $2 \, mm$ is illuminated by monochromatic light of wavelength $500 \, nm$. The distance between the first minima on either side on a screen at a distance of $1 \, m$ is ........ $mm$.

Angular width $(\beta)$ of the central maximum of a diffraction pattern on a single slit does not depend upon

In deriving the single slit diffraction pattern,it is stated that the intensity is zero at angles of $\theta = n\lambda / a$,where $a$ is the slit width. Justify this by suitably dividing the slit to bring out the cancellation.

$A$ 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 $.....$