$A$ screen is placed $50\,cm$ from a single slit,which is illuminated with $6000\,\mathring{A}$ light. If the distance between the first and third minima in the diffraction pattern is $3\,mm,$ the width of the slit is ......$mm$.

The phenomenon of diffraction of light was discovered by

$A$ single slit diffraction pattern is formed with white light. For what wavelength of light the $4^{\text{th}}$ secondary maximum in the diffraction pattern coincides with the $3^{\text{rd}}$ secondary maximum in the pattern of light of wavelength $\lambda$?

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.

In the diffraction pattern of a single slit,the width of the secondary maxima compared to the central maximum is:

When you look at a clear blue sky, you see tiny specks and hair-like structures floating in your view, called "floaters". This is basically