Object Distance for Coincident Image in Silvered Biconvex Lens

To find the required distance of the object from the silvered biconvex lens such that the image coincides with the object, we can treat the silvered lens system as an equivalent curved mirror.

Step 1: Focal Length of the Unsilvered Lens

The focal length $f_L$ of a thin biconvex lens in air is given by the Lens Maker's Formula: $\frac{1}{f_L} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

Given:

• Refractive index of glass, $\mu = 1.5$
• Radii of curvature: $R_1 = +20\text{ cm}$ and $R_2 = -20\text{ cm}$

Substituting these values: $\frac{1}{f_L} = (1.5 - 1)\left(\frac{1}{20} - \left(-\frac{1}{20}\right)\right) = 0.5 \times \frac{2}{20} = \frac{1}{20}\text{ cm}^{-1}$ $f_L = 20\text{ cm}$

Step 2: Focal Length of the Silvered Surface

The silvered surface acts as a concave mirror to the light travelling inside the glass medium. Its focal length $f_M$ is given by: $f_M = \frac{R}{2} = \frac{20}{2} = 10\text{ cm}$

Step 3: Equivalent Focal Length of the Silvered Lens

When light passes through the lens, reflects off the silvered back surface, and passes back through the lens, the equivalent power $P_{eq}$ of the system is the sum of the powers of two refractions through the lens and one reflection from the mirror: $\frac{1}{F_{eq}} = \frac{2}{f_L} + \frac{1}{f_M}$

Substituting $f_L = 20\text{ cm}$ and $f_M = 10\text{ cm}$: $\frac{1}{F_{eq}} = \frac{2}{20} + \frac{1}{10} = \frac{1}{10} + \frac{1}{10} = \frac{1}{5}\text{ cm}^{-1}$ $F_{eq} = 5\text{ cm}$

Thus, the system behaves as an equivalent concave mirror of focal length $F_{eq} = 5\text{ cm}$.

Step 4: Condition for Coincident Image and Object

For a mirror system, the image coincides with the object when the object is placed at its center of curvature ($u = 2 F_{eq}$): $u = 2 \times F_{eq} = 2 \times 5\text{ cm} = 10\text{ cm}$

Thus, the object should be placed at a distance of $10\text{ cm}$ from the lens.

Correct Option: A