Physics - Wave Optics Question with Solution | TestHub

The fringe width under initial condition, slit width $d$ and the distance between the slit and the eyepiece is $D$ : $X=\frac{\lambda D}{d}$ The fringe width under new situation, slit width $2d$ and the distance between the slit and the eyepiece is $2D$ : $\begin{array}{rl}& X^{\prime }=\frac{\lambda (2D)}{(2d)}\\ & \therefore X=X^{\prime }\end{array}$