In a Young's double slit experiment,light of $500 \ nm$ is used to produce an interference pattern. When the distance between the slits is $0.05 \ mm$,the angular width (in degrees) of the fringes formed on the distant screen is close to $........^o$.

Young's double-slit experiment is performed in air and then performed in water. How does the fringe width change?

The path difference between two identical light waves at a point $Q$ on the screen is $3 \mu m$. If the wavelength of the waves is $5000 \text{ Å}$,then at point $Q$ there is:

In Young's double-slit experiment,the intensity at a point on the screen is $I_0/4$. Find the angular position of this point.

In $Y.D.S.E.,$ the slits are $0.2 \, cm$ apart and are illuminated by yellow light $(\lambda = 600 \, nm).$ What would be the fringe width on a screen placed $1 \, m$ from the plane of the slits if the whole system is immersed in water of refractive index $\mu = \frac{4}{3}$ (in $, mm$)?

In Young's double-slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be: (in $: 1$)