Animated Solution for Physics - Optics: The size of the image of an object, which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image.

A Tale of Two Lenses

Imagine you are tracking the journey of light rays coming from a very distant object. Because the object is at infinity, the rays arriving at our first optical element—a convex lens—are perfectly parallel.

When these parallel rays pass through the convex lens, they are bent and converge exactly at its focal plane. Since the focal length of this convex lens is given as $30\text{ cm}$, the first image, which we will call $I_1$, is formed exactly $30\text{ cm}$ away from the lens. We are also given a crucial piece of information: the size of this first image $I_1$ is $2\text{ cm}$.

Now, let's introduce a twist to our optical setup. A concave lens is placed between the convex lens and the image $I_1$, specifically at a distance of $26\text{ cm}$ from the convex lens.

Because of this new obstacle, the converging rays will be intercepted before they can actually meet to form $I_1$. To the concave lens, these incoming converging rays appear to be heading towards a point behind it. Therefore, $I_1$ will now act as a virtual object for our concave lens.

The distance of this virtual object from the concave lens is simply the difference in their positions: $30\text{ cm}-26\text{ cm}=+4\text{ cm}$. According to the Cartesian sign convention, since the virtual object is in the direction of the incident light, the object distance $u$ is positive.

To find where the final image $I_2$ is formed, we apply the lens formula for the concave lens:

Here, $u=+4\text{ cm}$, and the focal length $f$ of the concave lens is $-20\text{ cm}$. Substituting these values into our equation:

Moving the $\frac{1}{4}$ to the right side, we get:

Taking the common denominator as $20$, this becomes:

So, $v=+5\text{ cm}$. This means the final image $I_2$ is formed $5\text{ cm}$ to the right of the concave lens.

Now that we have the object and image distances for the concave lens, we can find its magnification $m$:

Finally, we can determine the size of the new image. The new size $h_2$ is simply the magnification multiplied by the size of the virtual object $h_1$:

And there we have it! The final size of the image is $2.5\text{ cm}$. Notice how the concave lens, which usually diminishes real objects, actually magnified our virtual object here! This fascinating result occurs because the virtual object was placed between the optical center and the focus of the concave lens.