A thin convex lens of focal length 15 cm has a point object on its axis 25 cm to its left. A plane mirror, perpendicular to the axis, is placed 28 cm…

First pass: $\dfrac1v-\dfrac1{(-25)}=\dfrac1{15}$ gives $v_1=37.5\ \mathrm{cm}$ to the right of the lens, so the rays are converging towards a point that lies $9.5\ \mathrm{cm}$ beyond the mirror (mirror at $28\ \mathrm{cm}$). The plane mirror reflects the converging beam so that it forms a real image at the mirror image of this point: $2(28)-37.5=18.5\ \mathrm{cm}$ from the lens, between lens and mirror. Second pass: this point is a real object for the lens at $u=-18.5\ \mathrm{cm}$ (light now travels leftward, so use distances measured in that direction), with $f=15$: $\dfrac1v=\dfrac1{15}-\dfrac1{18.5}$, i.e. $v=\dfrac{15\times18.5}{3.5}=79.29\ \mathrm{cm}$. The final image is real, to the left of the lens. A slip is to take the object for the second pass as $37.5$ or to treat the mirror as moving the image to $37.5-28$.