Fringes will be observed in the region between P₁ and P₂ because the reflected rays lie only in this region.

$$\begin{gathered} \frac{\mathrm{BS}}{\mathrm{BD}}=\frac{{\mathrm{OP}}_2}{\mathrm{OD}} \\ \Rightarrow {\mathrm{OP}}_2=39\mathrm{mm} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{BS}}{\mathrm{BC}}=\frac{{\mathrm{OP}}_1}{\mathrm{OC}} \\ \Rightarrow {\mathrm{OP}}_1=19\mathrm{mm} \end{gathered}$$

Thus, ${\mathrm{P}}_1{\mathrm{P}}_2=39-19=20\mathrm{mm}$

Wavelength, $\mathrm{\lambda }=\frac{\mathrm{c}}{\mathrm{f}}=\frac{3\times {10}^8}{6\times {10}^{14}}=5\times {10}^{-7}\mathrm{m}$

Fringe width, $\mathrm{\beta }=\frac{\mathrm{\lambda D}}{\mathrm{d}}=\frac{5\times {10}^{-7}\times \left(5+5+190\right)\times {10}^{-2}}{\left(1+1\right)\times {10}^{-3}}=5\times {10}^{-4}\mathrm{m}$

Number of fringes, $\mathrm{n}=\frac{{\mathrm{P}}_1{\mathrm{P}}_2}{\mathrm{\beta }}=\frac{20\times {10}^{-3}}{5\times {10}^{-4}}=40=8\times 5$