Animated Solution for Physics - Optics: A thin plano-convex lens of focal length $f$ is split into two halves. One of the halves is shifted along the optical axis. The separation between object and image planes is $1.8\text{ m}$. The magnification of the image formed by one of the half lens is $2$. Find the focal length of the lens and separation between the halves. Draw the ray diagram for image formation.

A Journey Through the Displacement Method

Imagine taking a perfectly good plano-convex lens and slicing it right down the middle along its optical axis. What happens to its optical properties? The beautiful truth of optics is that each half continues to bend light exactly as it did before. The focal length $f$ remains completely unchanged. In this fascinating problem, we explore what happens when these two identical halves are separated and used to project an image onto the exact same screen.

We are given a fixed distance between the object and the image plane: $D=1.8\text{ m}$. We are also told that one of the half-lenses produces an image with a magnification of $2$.

Since the image is being projected onto a physical plane (a screen), it must be a real image. For a single thin lens, real images are always inverted, which means our magnification is actually $m=-2$.

Let's focus on this first half-lens. We know that the magnitude of magnification is the ratio of the image distance to the object distance:

We also know that the total distance is the sum of the magnitudes of the object and image distances:

Unlocking the Focal Length

By substituting our magnification relationship into the distance equation, we can easily solve for the exact positions:

This gives us an object distance magnitude of $|u_1|=0.6\text{ m}$ and an image distance magnitude of $|v_1|=1.2\text{ m}$. Applying the standard Cartesian sign convention, we have $u_1=-0.6\text{ m}$ and $v_1=+1.2\text{ m}$.

Now, we can deploy the trusty lens formula to uncover the focal length $f$:

Finding the Separation

Now for the twist: the second half of the lens is shifted by some distance $d$ along the optical axis, yet it still forms a sharp image on the exact same screen!

Because it has moved further from the object by a distance $d$, its new object distance magnitude is $|u_2|=0.6+d$. Consequently, its distance to the screen shrinks, making the new image distance magnitude $|v_2|=1.2-d$. Applying our sign convention, we get $u_2=-(0.6+d)$ and $v_2=+(1.2-d)$.

Let's feed these new coordinates back into the lens formula, knowing that the focal length is still $0.4\text{ m}$:

To solve this, we find a common denominator. Notice how beautifully the $d$ terms cancel out in the numerator:

Cross-multiplying yields a simple quadratic equation:

Since the lenses are physically separated, $d$ cannot be zero. Therefore, $2.5d=1.5$, which gives us our separation: $d=0.6\text{ m}$.

We have successfully deduced that the focal length is $0.4\text{ m}$ and the separation between the halves is $0.6\text{ m}$.

Did you know there's a beautiful shortcut here? This setup is a classic manifestation of the Displacement Method. The separation between the two lens positions $d$ that form an image on a fixed screen separated by distance $D$ is given by the elegant formula $d=\sqrt{D(D-4f)}$.

Let's plug in our values to verify: !

Physics is truly poetic when the mathematics aligns so perfectly with the physical reality.