Animated Solution for Physics - Optics: In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is $R$. Determine the focal length of this system.

The problem of finding the focal length of a lens system with multiple media might seem daunting at first glance. However, by breaking it down into atomic steps and applying the fundamental principles of optics, the solution reveals itself beautifully. Let's dive into the physics behind this meniscus lens!

Imagine you are a light ray traveling from the left. You first encounter a spherical surface separating medium ${\mu }_1$ and medium ${\mu }_2$. After passing through this first surface, you travel through the lens and hit a second spherical surface separating medium ${\mu }_2$ and medium ${\mu }_3$.

To find the focal length of this entire system, we need to determine where parallel rays coming from infinity will finally converge.

For any single spherical refracting surface, the relationship between the object distance $u$, the image distance $v$, and the radius of curvature $R$ is given by the master equation:

We will apply this equation sequentially to both surfaces.

Let's focus on the first surface on the left. The incident rays are parallel, which means our object is at infinity ($u=-\infty$). The light is traveling from medium ${\mu }_1$ into medium ${\mu }_2$.

By sign convention, since the surface bulges towards the left (towards the incident light), its center of curvature lies to the right. Therefore, the radius of curvature is positive ($+R$). Plugging these values into our master equation, we get:

Since any finite number divided by infinity is zero, the second term vanishes completely. This leaves us with the position of the first intermediate image:

This intermediate image at $v_1$ will now act as a virtual object for the second surface.

Now, let's analyze the second surface. The light is now traveling from medium ${\mu }_2$ into medium ${\mu }_3$. The object distance for this surface is the image distance from the first surface, so $u=v_1$.

Just like the first surface, this second surface also bulges to the left, meaning its center of curvature is also on the right. Thus, its radius of curvature is also positive ($+R$). Applying the master equation again:

We now have a system of two equations. Notice how the term $\frac{{\mu }_2}{v_1}$ appears in both equations, but with opposite signs. This is a massive hint! Let's add the two equations together to eliminate this intermediate term:

Because the denominators on the right side are the same, we can simply add the numerators. Watch what happens to the ${\mu }_2$ terms:

The ${\mu }_2$ terms beautifully cancel out, leaving us with:

Since the original incident rays were parallel, the final image distance $v_2$ is, by definition, the focal length $f$ of the entire lens system. Rearranging the equation to solve for $f$, we get our final, elegant result:

And there we have it! By systematically applying the single surface refraction formula, we've successfully navigated through multiple media to find the focal length.