Animated Solution for Physics - Optics: A spherical surface of radius of curvature $R$, separates air (refractive index $1.0$) from glass (refractive index $1.5$). The centre of curvature is in the glass. A point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $O$ and $PO=OQ$. The distance $PO$ is equal to

Imagine a spherical surface that acts as a boundary between two different media: air on one side and glass on the other. The problem tells us that the center of curvature of this surface lies inside the glass. This means if we are looking from the air, the glass surface bulges outwards towards us—it's a convex refracting surface.

We place a point object $P$ in the air. Light rays from $P$ travel through the air, hit the spherical surface, and refract into the glass to form a real image $Q$. The line connecting the object and the image passes through the pole $O$ of the spherical surface. We are given a beautiful symmetry: the distance from the pole to the object ($PO$) is exactly equal to the distance from the pole to the image ($OQ$). Let's call this unknown distance $X$.

To solve any problem involving refraction at a single spherical surface, we rely on the master equation:

Here, ${\mu }_1$ is the refractive index of the medium where the incident rays originate (air, so ${\mu }_1=1.0$), and ${\mu }_2$ is the refractive index of the medium where the refracted rays travel (glass, so ${\mu }_2=1.5$). The terms $u$, $v$, and $R$ represent the object distance, image distance, and radius of curvature, respectively.

The most critical step in optics is applying the Cartesian sign convention correctly. We take the pole $O$ as our origin $(0,0)$. The direction of incident light (from left to right) is taken as positive.

1. Object Distance ($u$): The object $P$ is in the air, to the left of the pole. Therefore, $u=-X$. 2. Image Distance ($v$): The real image $Q$ forms in the glass, to the right of the pole. Therefore, $v=+X$. 3. Radius of Curvature ($R$): The center of curvature is in the glass, to the right of the pole. Therefore, the radius is $+R$.

Now, let's substitute these values into our master equation:

Notice how the two negative signs in the second term cancel each other out:

Since the denominators on the left side are the same, we can simply add the numerators:

Finally, solving for $X$, we get:

Thus, the distance $PO$ is exactly $5R$.