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 ?

For a thin symmetrical convex lens each surface has radius of curvature $$R$$ and refractive index $$\mu$$ with respect to air.

Step 1 : Focal length of the glass lens
Lens-maker’s formula for a thin lens in air is
$$\frac{1}{f_{\text{lens}}} = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
For a biconvex lens $$R_1 = +R$$ (centre to the right) and $$R_2 = -R$$ (centre to the left).
Hence
$$\frac{1}{f_{\text{lens}}} = (\mu-1)\left(\frac{1}{R}-\frac{-1}{R}\right) = (\mu-1)\left(\frac{2}{R}\right)$$
Therefore
$$f_{\text{lens}} = \frac{R}{2(\mu-1)} \quad -(1)$$

Step 2 : Power of the lens
Optical power is $$P = \dfrac{1}{f}$$ (in dioptres when $$f$$ is in metres).
From $$(1)$$
$$P_{\text{lens}} = \frac{1}{f_{\text{lens}}} = \frac{2(\mu-1)}{R} \quad -(2)$$

Step 3 : Power of the silvered (concave) surface
After silvering the second surface acts as a concave mirror of radius $$R$$.
Focal length of a spherical mirror is $$f_{\text{mirror}} = \dfrac{R}{2}$$,
so its power is
$$P_{\text{mirror}} = \frac{1}{f_{\text{mirror}}} = \frac{2}{R} \quad -(3)$$

Step 4 : Equivalent focal length of the silvered lens
Light passes through the lens, reflects, and then passes through the lens again, so the lens power is used twice. The total power of the system is
$$P_{\text{eq}} = 2P_{\text{lens}} + P_{\text{mirror}}$$
Substituting from $$(2)$$ and $$(3)$$:
$$P_{\text{eq}} = 2\left(\frac{2(\mu-1)}{R}\right) + \frac{2}{R} = \frac{4(\mu-1) + 2}{R} = \frac{2(2\mu-1)}{R}$$
Hence the equivalent focal length $$F$$ of the silvered lens is
$$F = \frac{1}{P_{\text{eq}}} = \frac{R}{2(2\mu-1)} \quad -(4)$$

Step 5 : Condition for object and image to coincide
The silvered lens behaves like a concave mirror of focal length $$F$$. For a concave mirror, an object placed at its centre of curvature (distance $$2F$$) produces an image at the same position.
Therefore the required object distance from the lens is
$$u = 2F = 2\left(\frac{R}{2(2\mu-1)}\right) = \frac{R}{2\mu-1}$$

Answer : The object must be placed at a distance $$\displaystyle \frac{R}{2\mu-1}\;$$ from the lens (Option D).