In a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

The figure shows a schematic diagram of the arrangement of Young's Double Slit Experiment. If the distance $d$ is varied,then identify the correct statement.

The width of the fringe is $2\,mm$ on the screen in a double-slit experiment for light of wavelength $400\,nm$. The width of the fringe for light of wavelength $600\,nm$ will be $..............\,mm$.

In Young's experiment,the distance between slits is $0.28 \,mm$ and the distance between the slits and the screen is $1.4 \,m$. The distance between the central bright fringe and the third bright fringe is $0.9 \,cm$. What is the wavelength of the light used in $\mathring{A}$?

In which of the following is the interference due to the division of wave front?