$A$ single slit of width $a$ is illuminated by violet light of wavelength $400 \ nm$ and the width of the diffraction pattern is measured as $y.$ When half of the slit width is covered and illuminated by yellow light of wavelength $600 \ nm$,the width of the diffraction pattern is

$A$ light source with a wavelength of $5000 \, \mathring A$ produces a single-slit diffraction pattern. The first minimum in the diffraction pattern is observed at a distance of $5 \, mm$ from the central maximum. The distance between the slit and the screen is $2 \, m$. Find the width of the slit. (in $, mm$)

The angular width of the central maximum in a single-slit diffraction pattern does not depend on .........

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 ......... $nm$.

$A$ 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 $.....$

$A$ slit of width $a$ is illuminated by red light of wavelength $6500 \, \mathring{A}$. The first minimum will fall at $\theta = 30^{\circ}$ if $a$ is equal to: