Animated Solution for Physics - Optics: A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature $R=14\text{ cm}$. For this bi-convex lens, for an object distance of $40\text{ cm}$, the image distance will be

This problem is a beautiful demonstration of how complex optical systems can be broken down into simpler, manageable parts. We are given a bi-convex lens, but it's not a standard uniform lens. It is constructed by joining two distinct plano-convex lenses, each made of a material with a different refractive index. Let's embark on a journey to find where this composite lens forms an image.

Imagine you are holding this composite lens. The left half is a plano-convex lens with a refractive index $n_1=1.5$. Its first surface (facing the object) is convex, and its second surface (the interface) is perfectly flat. The right half is another plano-convex lens with a refractive index $n_2=1.2$. Its first surface is the flat interface, and its second surface is convex.

Both curved surfaces share the same radius of curvature, $R=14\text{ cm}$. Our goal is to find the image distance $v$ for an object placed at $u=-40\text{ cm}$. To do this, we first need to determine the equivalent focal length $F$ of the entire system.

When two thin lenses are placed in contact, their powers add up. This means their equivalent focal length $F$ is given by the reciprocal sum of their individual focal lengths:

To find $f_1$ and $f_2$, we rely on the trusty Lens Maker's Formula:

Let's focus on the first lens ($L_1$). The light hits the convex surface first. According to the Cartesian sign convention, the center of curvature for this surface lies to the right (in the direction of light), so $R_1=+14\text{ cm}$. The second surface is plane, which means it's essentially a sphere with an infinite radius, so $R_2=\infty$.

Substituting these into the Lens Maker's Formula:

Since $\frac{1}{\infty }=0$, this simplifies beautifully:

Now, let's analyze the second lens ($L_2$). The light first encounters the plane interface, so $R_1=\infty$. The light then exits through the convex surface. The center of curvature for this surface lies to the left (opposite to the direction of light), making $R_2=-14\text{ cm}$.

Watch out for the double negative! It becomes positive:

Now that we have the optical power of both components, we simply add them to find the power of the composite lens:

To add these fractions, we find a common denominator, which is $140$:

This tells us that the equivalent focal length of our bi-convex lens is $F=+20\text{ cm}$. The positive sign confirms that the combination acts as a converging lens.

With the equivalent focal length in hand, we can now determine the image position using the standard thin lens formula:

We are given the object distance $u=-40\text{ cm}$. Substituting our known values:

Isolating $\frac{1}{v}$:

Therefore, the image distance is $v=+40\text{ cm}$. The positive sign indicates that a real image is formed $40\text{ cm}$ to the right of the lens. This elegant result showcases how fundamental principles seamlessly combine to solve complex optical setups.