$A$ wavefront presents one,two and three $HPZ$ at points $A, B$ and $C$ respectively. If the ratio of consecutive amplitudes of $HPZ$ is $4 : 3$,then the ratio of resultant intensities at these points will be

Assertion : When a tiny circular obstacle is placed in the path of light from some distance,a bright spot is seen at the centre of the shadow of the obstacle.
Reason : Destructive interference occurs at the centre of the shadow.

$A$ 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 $.....$

Light of wavelength $5000 \, \mathring{A}$ is incident on a single slit such that the first minima is formed at a distance $5 \, mm$ from the centre. If the screen is placed $2 \, m$ away,then find the width of the slit in $mm$.

In a single slit diffraction pattern,a light of wavelength $6000 \mathring A$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3 \text{ mm}$ when the screen is placed $50 \text{ cm}$ away from the slit. The width of the slit is . . . . . . $\times 10^{-4} \text{ m}$.

$A$ slit of width $a$ is illuminated by red light of wavelength $6500 Å$. If the first diffraction minimum falls at $30^{\circ}$,then the value of $a$ is