Animated Solution for Physics - Optics: A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object

When a Lens Becomes a Mirror

Imagine taking a standard plano-convex lens and coating its curved surface with a highly reflective layer of silver. What happens to light when it enters this system? It doesn't just pass through; it embarks on a fascinating journey. The light first refracts as it enters the plane surface, then it reflects off the silvered curved surface, and finally, it refracts one last time as it exits the plane surface.

Because the light is ultimately reflected back, this entire optical combination behaves exactly like a single spherical mirror. But how do we find the focal length of this equivalent mirror?

Combining Powers

To find the focal length of this equivalent mirror, we use the power combination formula. The total power of the system is simply the sum of the powers of each individual optical event:

Let's break this down. First, we calculate the power of the unsilvered plano-convex lens ($P_L$). Using the lens maker's formula, we know that the first surface is plane ($R_1=\infty$) and the second surface is convex towards the inside ($R_2=-R$). Plugging these into the formula gives:

Next, we look at the silvered curved surface. It acts as a concave mirror with a radius of curvature $R$. Its focal length is $f_M=-\frac{R}{2}$. Since the power of a mirror is defined as $P_M=-\frac{1}{f_M}$, the mirror's power becomes:

Now, we substitute the lens power and the mirror power back into our master equation:

Notice how beautifully the algebra simplifies. The $-2$ and $+2$ cancel out perfectly, leaving us with:

Since the equivalent focal length is $F_{eq}=-\frac{1}{P_{eq}}$, we get:

Now, we bring in the given values: $R=30\text{ cm}$ and $\mu =1.5$. Calculating this gives us an equivalent focal length of:

So, our complex silvered lens system is mathematically identical to a simple concave mirror with a focal length of $10\text{ cm}$!

The question asks for the object distance required to form a real image of the exact same size as the object. For a concave mirror, this specific condition is met only when the object is placed exactly at the center of curvature, which is twice the focal length ($2f$).