Animated Solution for Physics - Optics: A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let $O$ be the pole of the mirror and $C$ its centre of curvature. A point object is placed at $C$. It has a real image, also located at $C$. If the mirror is now filled with water, the image will be

A Simple Reflection Imagine a concave mirror placed horizontally on a table. When an object is placed exactly at its center of curvature, $C$, the light rays strike the mirror normally. Because they hit at a $90^{\circ }$ angle to the surface, they retrace their exact path backwards, forming a real image precisely at $C$. This is a classic property of spherical mirrors.

The Birth of a Composite System Now, let's make things interesting by filling the mirror with water. The system is no longer just a mirror; it has transformed into a composite optical system. The water forms a plano-convex lens sitting right on top of the concave mirror.

When light travels from the object at $C$, it now undergoes three distinct optical events: 1. It refracts as it enters the water lens. 2. It reflects off the concave mirror at the bottom. 3. It refracts again as it exits the water lens.

To analyze this, we can treat the entire setup as a single equivalent mirror. The equivalent power of this system is the sum of the powers of its individual components: $P_{eq}=P_L+P_M+P_L=2P_L+P_M$.

First, the plano-convex water lens. Using the lens maker's formula, the power $P_L$ is given by: Since the top surface is flat ($R_1=\infty$) and the bottom surface matches the mirror ($R_2=-R$), this simplifies to:

Next, the power of the concave mirror, $P_M$, is defined as $-1/f_M$. Since the focal length of a concave mirror is $-R/2$, we get:

Adding these together gives us the equivalent power of the system:

The equivalent focal length $F$ is simply the negative reciprocal of the equivalent power:

Substituting this into our focal length expression yields:

Now, we can use the standard mirror formula to find the new image position. The object is still at $C$, so the object distance $u=-R$.

Solving for $v$:

Conclusion: The negative sign indicates that the image is real (formed in front of the mirror). Since the magnitude $0.6R$ is less than the radius of curvature $R$, the image is located exactly between the pole $O$ and the center of curvature $C$.