The cause of Fraunhofer lines is

In a Fraunhofer diffraction pattern,the slit width is $0.2 \ mm$ and the screen is at $2 \ m$ away from the lens. If the distance between the first minimum on either side of the central maximum is $1 \ cm$,the wavelength of light used is: (in $Å$)

In a Fraunhofer diffraction,light of wavelength $\lambda$ is incident on a slit of width $d$. The diffraction pattern is observed on a screen placed at a distance $D$. If the linear width of the central maximum is equal to two times the width of the slit,then the value of $D$ is:

The condition for observing Fraunhofer diffraction from a single slit is that the wave front incident on the slit should be ............

Two towers on top of two hills are $40 \, km$ apart. The line joining them passes $50 \, m$ above a hill halfway between the towers. What is the longest wavelength of radio waves (in $cm$),which can be sent between the towers without appreciable diffraction effects?

In the Fraunhofer diffraction experiment, $L$ is the distance between the screen and the obstacle, $b$ is the size of the obstacle, and $\lambda$ is the wavelength of the incident light. The general condition for the applicability of Fraunhofer diffraction is: