Given is a thin convex lens of glass (refractive index mu) and each side having radius of curvature R. One side is polished for complete reflection. At what distance from the lens, an object be placed on the optic axis so that the image gets formed on the object itself. (1) R/mu (2) R/(2mu–3) (3) muR (4) R/(2mu–1)

1. Know the individual players

1. Thin Convex Lens
   Lens–maker’s formula for equal radii $R$ on both sides:

So

2. Silvered Surface (Spherical Mirror)
   Radius of curvature $=R$ (because the back face of the lens is silvered).
   For a spherical mirror:

Its power (ray-matrix language) is $-\dfrac{2}{R}$.

2. Optical path of light

Object → (distance $u$) → Lens → Mirror → back through same Lens → comes out → (distance $u$) → back to object plane.

3. Matrix (ABC-D) approach

1. Translation (distance $d$): $\begin{bmatrix}1 & d\\ 0 & 1\end{bmatrix}$
2. Thin lens (focal length $f$): $\begin{bmatrix}1 & 0\\ -\tfrac{1}{f} & 1\end{bmatrix}$
3. Spherical mirror ($R$): $\begin{bmatrix}1 & 0\\ -\tfrac{2}{R} & 1\end{bmatrix}$

Sequence:
$P(u)$ → $B$ → $M$ → $B$ → $P(u)$
Multiply them and demand the upper-right element $(B)$ of the final matrix to be zero (that condition gives image exactly on object plane).

This gives

4. Substitute $f$

Hence

So option (4) is correct.

🔍 What is happening?

Imagine a magnifying glass (a convex lens) whose back surface has been painted with silver just like a mirror. So, light first goes through the glass, hits the silvered back, bounces, comes back again through the same glass and finally comes out.

We want to put a small candle in front of this shiny magnifying glass in such a way that after all those trips the final picture of the candle falls exactly on the candle itself. It is like throwing a ball at a wall and wanting it to come back to the exact spot where you are standing.

The secret is to find the correct distance of the candle from the lens so the lens–mirror combo behaves perfectly like a boomerang for light.

Step-by-Step Calculation

1. Lens focal length

2. ABCD matrices Translation $P(u)=\begin{bmatrix}1 & u\\ 0 & 1\end{bmatrix}$
   Lens $B=\begin{bmatrix}1 & 0\\ -\tfrac{1}{f} & 1\end{bmatrix}$
   Mirror $M=\begin{bmatrix}1 & 0\\ -\tfrac{2}{R} & 1\end{bmatrix}$

Total matrix

After multiplication (skipped algebra shown in theory section):

3. Self-imaging condition
   We require $B=0$.

4. Plug $f$ and solve for $u$

Final Answer