A thin symmetrical double convex lens of refractive index μ_2=1.5...

For first refracting surface, For parallel beam of light; $-\frac{\mu_1}{u}+\frac{\mu_2}{V}=\frac{\mu_2-1}{R}$$\begin{aligned} & \Rightarrow-\frac{1 \cdot 4}{-\infty}+\frac{1 \cdot 5}{V}=\frac{1 \cdot 5-1.4}{R} \\ & V=15 R\end{aligned}$$$For second refracting surface,$$\begin{aligned} & -\frac{\mu_2}{1.5 R}+\frac{\mu_3}{V^{\prime}}=\frac{\mu_3-\mu_2}{R} \\ & \Rightarrow-\frac{1.5}{15 R}+\frac{1.6}{V^{\prime}}=\frac{1.6-1.5}{R} \Rightarrow V^{\prime}=\infty\end{aligned}$$ Hence, the combination behaves as glass plate.