In Young's double slit experiment the distance between the slits...

In Young's double slit experiment fringe width $\beta=\frac{\lambda D}{d}$where $D=$ distance between slit and screen $d=$ distance between slit According to the question $D$ is double $\rightarrow 2 D$ and $d$ is half $\rightarrow\frac{d}{2}$ Ther New fringe wilth will be $\frac{\lambda \times 2 D}{\frac{d}{2}}=\frac{4 \lambda D}{d}=4\left(\frac{\lambda D}{d}\right)=\beta\beta^1=4 \beta$ i.e., becomes four times