A plano-convex glass lens (μ=1.5) has focal length 0.50 m in air. Its convex surface is placed on a flat glass plate of the same material, and the…

First find the radius of curvature from the lens-maker formula for a plano-convex lens: $\frac1f=(\mu-1)\frac1R$, so $R=0.5\times0.5=0.25$ m. The film is liquid of index $1.2$, which is less than that of glass. Reflection at the glass-to-liquid boundary has no phase change, while that at the liquid-to-glass boundary has a change of $\pi$, so the central spot is dark and dark rings satisfy $2\mu t=n\lambda$. For a gap $t=\frac{r^2}{2R}$ this gives $r_n^2=\frac{n\lambda R}{\mu}$. For $n=5$: $r^2=\frac{5\times600\times10^{-9}\times0.25}{1.2}=6.25\times10^{-7}\ \mathrm{m^2}$, so $r=0.79$ mm. Omitting the liquid index gives $0.87$ mm, using a bright-ring condition gives $0.75$ mm, and multiplying instead of dividing by $\mu$ gives $0.95$ mm.