A biconvex thick lens is made of glass ( μ = 1.50 ). Both of the...

For the first surface, using the formula for refraction at a spherical surface:

$\dfrac{\mu_2}{v_1} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R_1}$

Given $\mu_1 = 1$, $\mu_2 = 1.5$, $u = -\infty$, and $R_1 = +10 \text{ cm}$:

$\dfrac{1.5}{v_1} - 0 = \dfrac{1.5 - 1}{10}$

$\dfrac{1.5}{v_1} = \dfrac{0.5}{10} \Rightarrow v_1 = 30 \text{ cm}$

The image formed by the first surface acts as a virtual object for the second surface. The object distance for the second surface is:

$u_2 = v_1 - t = 30 - 5 = +25 \text{ cm}$

For the second surface, light travels from glass to air. Using $\mu_1 = 1.5$, $\mu_2 = 1$, and $R_2 = -10 \text{ cm}$:

$\dfrac{\mu_2}{v_2} - \dfrac{\mu_1}{u_2} = \dfrac{\mu_2 - \mu_1}{R_2}$

$\dfrac{1}{v_2} - \dfrac{1.5}{25} = \dfrac{1 - 1.5}{-10}$

$\dfrac{1}{v_2} - \dfrac{3}{50} = \dfrac{-0.5}{-10} = \dfrac{1}{20}$

$\dfrac{1}{v_2} = \dfrac{1}{20} + \dfrac{3}{50} = \dfrac{5 + 6}{100} = \dfrac{11}{100}$

$v_2 = \dfrac{100}{11} \approx 9.1 \text{ cm}$

The final image is formed at $9.1 \text{ cm}$ from the farther surface on the other side of the lens.