Let radius of hemisphere is R. Initially flat surface will not effect the ray in this case because rays falls normally. So far curved surface $\frac{1}{(8-\mathrm{R})}=\frac{\mathrm{n}-1}{\mathrm{R}}$                                         …..(1)

Let after the hemisphere is turned around distance of image from flat surface is X, for curved surface, $\frac{\mathrm{n}}{{\mathrm{v}}^{\text{'}}}-\frac{1}{\mathrm{\infty }}=\frac{\mathrm{n}-1}{\left(\mathrm{R}\right)}\Rightarrow {\mathrm{v}}^{\text{'}}\frac{\mathrm{nR}}{\mathrm{n}-1}$

Now image by curved surface will be object for plane surface. So for plane surface

$\mathrm{u}={\mathrm{v}}^{\text{'}}-\mathrm{R}=\frac{\mathrm{nR}}{\mathrm{n}-1}-\mathrm{R}=\frac{\mathrm{nR}}{\mathrm{n}-1}$

for plane surface

$\frac{1}{\mathrm{v}}-\frac{\mathrm{n}}{\mathrm{u}}=\frac{1}{\mathrm{\infty }}$                ....(2)

$\Rightarrow \mathrm{v}=\frac{\mathrm{R}}{\mathrm{n}(\mathrm{n}-1)}$             ....(3)

From equation (1) and (2), we get

$\mathrm{v}=\frac{8}{{\mathrm{n}}^2}=\frac{8}{4}=2\mathrm{cm}$