Light of wavelength $5000 \, \text{Å}$ is incident on a slit of width $0.1 \, \text{mm}$. Find the width of the central bright fringe on a screen placed at a distance of $2 \, \text{m}$ in $\text{mm}$.

$A$ 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 $d$ is illuminated by white light. The first minimum for red light $(\lambda = 6500\,\mathring{A})$ will fall at $\theta = 30^\circ$ when $d$ will be

$A$ parallel beam of monochromatic light falls normally on a single narrow slit. The angular width of the central maximum in the resulting diffraction pattern

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 $D=$

To observe diffraction,the size of the obstacle