In Young's double slit experiment,the width of fringes can be increased if we decrease the

In Young's double-slit experiment,if a metal plate of thickness $t$ is placed in the path of one of the rays,then:

In an interference experiment, the $m^{th}$ bright fringe for light of wavelength $\lambda_1$ coincides with the $n^{th}$ dark fringe for light of wavelength $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ 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 $\lambda = \dots \mathring{A}$.

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width is

When the distance between the slits is increased $4$ times in $YDSE$,the fringe width will