In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of `n lambda/a `.Justify this by suitably dividing the slit to bring out the cancellation .

In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?

In the diffraction pattern due to a single slit of width 'd' with incident light of wavelength 'lambda' , at an angle of diffraction 'theta' , the condition for first minimum is

Describe with a neat diagram, the Fraunhofer's diffracting pattern due to a single slit.

In a single slit diffraction pattern, the distance between the 1st minima on the right and the 1st minima on the left is 5.2 mm. The screen on which the pattern is displayed is 80 cm from the slit. If lambda= 5460 overset@A . Calculate the slit width.

The angular width of the central maximum in a single slit diffraction pattern is 60^@ . The width of the slit is 1 mu m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the centres of each slit.)

Compare Young's Double slit interfemce pattern and single slit Diffraction pattern.

A lens of focal length f gives diffraction pattern of Fraunhoffer type of a slit having width a. If wavelength of light is lambda , the distance of first dark band and next bright band from axis is given by

Derive the condition for bright and dark fringes produced due to diffraction by a single slit .