$A$ microwave of wavelength $2.0 \ cm$ falls normally on a slit of width $4.0 \ cm$. The angular spread of the central maxima of the diffraction pattern obtained on a screen $1.5 \ m$ away from the slit,will be: (in $^{\circ}$)

$A$ monochromatic light is incident on a single slit of width $0.014 \ mm$. The angular position of the second bright line observed is $2.81^{\circ}$. Then the wavelength of the incident light is $\left[\sin \left(2.81^{\circ}\right)=0.049072\right]$ (in $Å$)

$A$ screen is placed at a certain distance from a narrow slit,which is illuminated by a parallel beam of monochromatic light. What will you observe if you scan the screen with the help of a microscope?

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 $D=$

For a parallel beam of monochromatic light of wavelength $\lambda$,diffraction is produced by a single slit whose width $a$ is of the order of the wavelength of the light. If $D$ is the distance of the screen from the slit,the width of the central maxima will be

In a single-slit diffraction pattern,if the source of light is replaced by a source of shorter wavelength,the width of the central maximum will: