Animated Solution for Physics - Optics: A point object is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm. The plane surface of the lens is now silvered. The image created by the system is at

A silvered lens is one of those classic JEE traps that makes you pause and ask: "Wait, is this a lens or a mirror?" The beautiful answer is that it's both! When light enters a silvered lens, it undergoes a fascinating three-part journey. Let's break down this optical adventure step-by-step, and then I'll show you a ninja technique to solve it in seconds.

The First Refraction Imagine a light ray starting from our object $O$, placed $20\text{ cm}$ in front of the lens. The first thing it encounters is the curved surface of the plano-convex lens. It refracts! We can find where these refracted rays are heading by applying the standard lens formula:

Substituting our known values ($u=-20\text{ cm}$ and $f=+15\text{ cm}$):

Solving this gives $v_1=+60\text{ cm}$. This means the lens is trying to converge the rays to a point $60\text{ cm}$ to its right. This is our first virtual image, $I_1$.

The Reflection But the rays never actually reach $I_1$! Before they can exit the lens, they hit the silvered back surface. This surface acts as a perfect plane mirror.

For a plane mirror, the image is formed at the exact same distance behind it as the object is in front. Our virtual object $I_1$ is at $+60\text{ cm}$, so the mirror reflects these rays, making them converge towards a new image, $I_2$, at $-60\text{ cm}$ (which is $60\text{ cm}$ to the left of the lens).

The Final Refraction Now for the final act. The reflected rays are travelling backwards (from right to left) and must exit the lens. They undergo a second refraction.

Here is the catch: Since the light is now travelling from right to left, we must take this new direction as our positive axis. The rays are converging towards $I_2$, which is $60\text{ cm}$ in this new positive direction. So, our new object distance is $u_3=+60\text{ cm}$.

Plugging this into the lens formula one last time:

This gives $v_3=+12\text{ cm}$. Since it's positive, it lies in the direction of the light, meaning the final image is formed $12\text{ cm}$ to the left of the system.

Equivalent Power Tracking the ray step-by-step builds great intuition, but during an exam, speed is key. A silvered lens can be treated as a single equivalent mirror! The total power of the system is the sum of the powers of each event:

The power of the lens is $P_L=\frac{1}{f_L}=\frac{1}{15}$. The power of a plane mirror is zero ($P_M=0$).

The equivalent focal length of this mirror is $F_{eq}=-\frac{1}{P_{eq}}=-7.5\text{ cm}$. The system behaves exactly like a concave mirror! Now, just apply the mirror formula once:

Solving this yields $v=-12\text{ cm}$. The negative sign confirms the image is formed in front of the mirror, $12\text{ cm}$ to the left. Boom! Same result, fraction of the time.