In a single slit diffraction experiment,when the distance of separation between the slit and the screen is doubled,the angular separation between the fringes

In a diffraction pattern due to a single slit of width $a$,the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \; \mathring{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

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.

$A$ narrow slit of width $2 \ mm$ is illuminated with a monochromatic light of wavelength $500 \ nm$. If the distance between the slit and the screen is $1 \ m$,then the first minima are separated by a distance of: (in $mm$)

In a single slit diffraction experiment, a parallel beam of green light of wavelength $\lambda = 550 \,nm$ passes through a slit of width $a = 0.20 \,mm$. The transmitted light is collected on a screen at a distance $D = 100 \,cm$ away. The distance of the first-order minima from the central maximum is $x \times 10^{-5} \,m$. The value of $x$ is:

$A$ 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