JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Geometrical Optics

Lens ($f=+20,u=-40$): $\frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}\Rightarrow v=+40$ cm, $m_1=-1$. This image lies $60-40=20$ cm in front of the mirror. Mirror ($f=10,u=-20$): $v=-20$ cm, $m_2=-1$; the image is $40$ cm from the lens. Light now returns through the lens ($u=-40,f=+20$): $v=+40$ cm, $m_3=-1$. Net magnification $(-1{)}^3=-1$, so the final image is $40$ cm from the lens (object side), same size, inverted. No option gives '$40$ cm from the lens, same size', so the question is a bonus.