A converging lens of focal length 15 cm and a converging mirror of...

For the first refraction through the converging lens:

Object distance, $u_1 = -30 \text{ cm}$

Focal length, $f_1 = +15 \text{ cm}$

Using the lens formula, $\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}$

$\dfrac{1}{v_1} - \dfrac{1}{-30} = \dfrac{1}{15} \Rightarrow v_1 = +30 \text{ cm}$

Magnification, $m_1 = \dfrac{v_1}{u_1} = \dfrac{30}{-30} = -1$

The first image is formed $30 \text{ cm}$ to the right of the lens.

For the reflection from the converging (concave) mirror:

Distance between the lens and the mirror is $50 \text{ cm}$.

Object distance for the mirror, $u_2 = -(50 - 30) = -20 \text{ cm}$

Focal length of the mirror, $f_2 = -10 \text{ cm}$

Using the mirror formula, $\dfrac{1}{v_2} + \dfrac{1}{u_2} = \dfrac{1}{f_2}$

$\dfrac{1}{v_2} + \dfrac{1}{-20} = \dfrac{1}{-10} \Rightarrow v_2 = -20 \text{ cm}$

Magnification, $m_2 = -\dfrac{v_2}{u_2} = -\dfrac{-20}{-20} = -1$

The second image is formed $20 \text{ cm}$ in front of the mirror, which is $30 \text{ cm}$ to the right of the lens. This coincides with the center of curvature of the mirror, so the rays retrace their path.

For the second refraction through the lens:

The light rays travel back towards the lens.

Object distance for the lens, $u_3 = -30 \text{ cm}$

Using the lens formula, $\dfrac{1}{v_3} - \dfrac{1}{u_3} = \dfrac{1}{f_1}$

$\dfrac{1}{v_3} - \dfrac{1}{-30} = \dfrac{1}{15} \Rightarrow v_3 = +30 \text{ cm}$

The final image is formed $30 \text{ cm}$ to the left of the lens, which is exactly at the original object position.

Magnification, $m_3 = \dfrac{v_3}{u_3} = \dfrac{30}{-30} = -1$

Total magnification, $m = m_1 \times m_2 \times m_3 = (-1) \times (-1) \times (-1) = -1$

Size of the final image $= |m| \times h_o = 1 \times 2.0 \text{ cm} = 2.0 \text{ cm}$.