$A$ point source is placed $20 \,cm$ to the left of a concave lens of focal length $10 \,cm$.
$(a)$ Where is the image formed?
$(b)$ Where to the right of the lens would you place a concave mirror of focal length $5 \,cm$,so that the final image is coincident with the source?
$(c)$ Where would the final image be formed,if the concave mirror is replaced by a plane mirror at the same position?

(D) Given: $u = -20 \,cm$,$f = -10 \,cm$ for the concave lens.
$(a)$ Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-10} \Rightarrow \frac{1}{v} = -\frac{1}{10} - \frac{1}{20} = -\frac{3}{20} \Rightarrow v = -\frac{20}{3} \,cm$.
The image is virtual and formed $6.67 \,cm$ to the left of the lens.
$(b)$ Let the mirror be at distance $x$ from the lens. The image formed by the lens acts as an object for the mirror. For the final image to coincide with the source,the rays must strike the mirror normally. This happens if the rays are directed towards the center of curvature of the mirror. The distance of the image from the mirror is $d = x + \frac{20}{3}$. For the rays to reflect back along the same path,this distance must equal the radius of curvature $R = 2|f_m| = 2 \times 5 = 10 \,cm$.
$x + \frac{20}{3} = 10 \Rightarrow x = 10 - 6.67 = 3.33 \,cm$.
$(c)$ If replaced by a plane mirror at $x = 3.33 \,cm$,the object distance for the mirror is $u_m = -(x + \frac{20}{3}) = -(3.33 + 6.67) = -10 \,cm$. The plane mirror forms an image at $v_m = +10 \,cm$ behind the mirror. This image acts as a virtual object for the lens at distance $u' = +(10 - 3.33) = +6.67 \,cm$. Using the lens formula: $\frac{1}{v'} - \frac{1}{6.67} = \frac{1}{-10} \Rightarrow \frac{1}{v'} = \frac{1}{6.67} - \frac{1}{10} = \frac{1}{20/3} - \frac{1}{10} = \frac{3}{20} - \frac{2}{20} = \frac{1}{20} \Rightarrow v' = +20 \,cm$.
The final image is formed $20 \,cm$ to the right of the lens.