A biconvex thin lens, fabricated from a material of refractive...

For the first refraction at the surface separating medium $\mu_3$ and the lens $\mu_2$, the incident light is parallel to the principal axis, so the object is at infinity ($u = -\infty$). The radius of curvature for the first biconvex surface is $R_1 = +R$.

Using the formula for refraction at a spherical surface:

$\dfrac{\mu_2}{v_1} - \dfrac{\mu_3}{-\infty} = \dfrac{\mu_2 - \mu_3}{R}$

$\Rightarrow \dfrac{\mu_2}{v_1} = \dfrac{\mu_2 - \mu_3}{R}$

For the second refraction at the surface separating the lens $\mu_2$ and medium $\mu_1$, the image formed by the first surface acts as a virtual object. The radius of curvature for the second biconvex surface is $R_2 = -R$.

$\dfrac{\mu_1}{v} - \dfrac{\mu_2}{v_1} = \dfrac{\mu_1 - \mu_2}{-R}$

$\Rightarrow \dfrac{\mu_1}{v} - \dfrac{\mu_2}{v_1} = \dfrac{\mu_2 - \mu_1}{R}$

Adding the two equations to eliminate $v_1$, we get:

$\dfrac{\mu_1}{v} = \dfrac{\mu_2 - \mu_3}{R} + \dfrac{\mu_2 - \mu_1}{R}$

$\Rightarrow \dfrac{\mu_1}{v} = \dfrac{2\mu_2 - \mu_1 - \mu_3}{R}$

$\Rightarrow v = \dfrac{\mu_1 R}{2\mu_2 - \mu_1 - \mu_3}$