$A$ diffraction pattern is obtained using a beam of red light. If the red light is replaced by blue light,then:

$A$ parallel beam of light of wavelength $500\; nm$ falls on a narrow slit and the resulting diffraction pattern is observed on a screen $1\;m$ away. It is observed that the first minimum is at a distance of $2.5\;mm$ from the centre of the screen. Find the width of the slit. (in $; mm$)

$A$ parallel beam of light of wavelength $\lambda$ is incident normally on a single slit of width $d$. Diffraction bands are obtained on a screen placed at a distance $D$ from the slit. The second dark band from the central bright band will be at a distance given by

In Fraunhofer diffraction due to a single slit,the direction of the second secondary maximum is given by ....... ($a$ is the width of the slit).

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.

$A$ circular disc is placed in front of a narrow source. When the point of observation is $2 \, m$ from the disc,it covers the first $HPZ$. The intensity at this point is $I$. When the point of observation is $25 \, cm$ from the disc,the intensity will be: