Animated Solution for Physics - Optics: An extended object is placed at point O, 10 cm in front of a convex lens ${\text{L}}_1$ and a concave lens ${\text{L}}_2$ is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is

Imagine you are standing in front of a complex optical system. We have an extended object placed $10\text{ cm}$ in front of a convex lens, $L_1$. But the journey of light doesn't end there! Just $10\text{ cm}$ behind $L_1$, we have a concave lens, $L_2$. Our ultimate goal is to find the total magnification of this entire dual-lens system. To do this, we must track the light rays step-by-step, treating the image formed by the first lens as the object for the second.

The Lens Maker's Formula

Before we can trace any rays, we need to determine the focal lengths of both lenses. We will use the Lens Maker's formula:

For the convex lens $L_1$, the refractive index $\mu$ is $1.5$, and the radii of curvature are $20\text{ cm}$. Following the Cartesian sign convention, the first surface is convex towards the object, making $R_1=+20\text{ cm}$, while the second surface curves inwards, making $R_2=-20\text{ cm}$.

Substituting these values, we get:

So, the focal length of our convex lens $L_1$ is exactly $+20\text{ cm}$.

Now, let's do the same for the concave lens $L_2$. Here, the first surface curves inwards, so $R_1=-20\text{ cm}$, and $R_2=+20\text{ cm}$. Substituting these into the formula gives us:

This yields a focal length of $-20\text{ cm}$. The negative sign perfectly matches the diverging nature of a concave lens.

Refraction at the Convex Lens

Alright, let's start the optical journey! The object is placed at $u_1=-10\text{ cm}$ from $L_1$. We'll use the standard thin lens formula to find where the first lens tries to form the image:

Substituting $u_1=-10\text{ cm}$ and $f_1=+20\text{ cm}$:

Moving the term to the right side, we get:

This means the first image, $I_1$, is formed at $v_1=-20\text{ cm}$. It is a virtual image, formed on the same side as the object!

What about the size of this intermediate image? The magnification $m_1$ is simply $v_1/u_1$:

The positive sign tells us the image is erect, and the magnitude of $2$ means it is twice as tall as our original object.

A Virtual Object Appears

Here is where the magic happens. This virtual image $I_1$ now acts as the object for our second lens, $L_2$. But we must measure its distance from $L_2$!

Since $L_2$ is $10\text{ cm}$ to the right of $L_1$, and $I_1$ is $20\text{ cm}$ to the left of $L_1$, the total object distance $u_2$ for the second lens is the sum of these distances, measured in the negative direction:

Refraction at the Concave Lens

Let's apply the lens formula one last time for $L_2$. We have $u_2=-30\text{ cm}$ and $f_2=-20\text{ cm}$:

Solving this carefully:

We find $v_2=-12\text{ cm}$. The final image is also virtual, formed $12\text{ cm}$ in front of the concave lens.

Let's find the magnification produced by the second lens alone:

Total Magnification

We are at the finish line! The total magnification of a multi-lens system is simply the product of their individual magnifications.

The final image is erect and exactly 80% the size of the original object. The correct option is (B).