Animated Solution for Physics - Optics: Formation of real image using a biconvex lens is shown below. If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen?

A Perfect Real Image

Imagine a classic optics experiment: a biconvex lens is placed perfectly between an object and a screen. The object is positioned exactly at a distance of $2f$ from the optical center of the lens. According to the principles of ray optics, the lens bends the incoming light rays to converge perfectly on the other side, forming a real, inverted image exactly at a distance of $2f$.

The screen captures this sharp image beautifully. Everything is in perfect harmony. But what happens if we disrupt this harmony by submerging the entire apparatus into water?

Enter the Lens Maker's Formula

When we immerse the lens in water, we aren't changing the physical shape of the lens—its radii of curvature, $R_1$ and $R_2$, remain exactly the same. However, we are fundamentally altering the environment in which the light travels.

To understand the impact of this change, we must consult the Lens Maker's Formula:

Here, ${\mu }_{rel}$ is the relative refractive index of the lens material with respect to its surrounding medium. In air, this is simply the refractive index of glass ($n_g=1.5$). But in water, it becomes the refractive index of glass with respect to water ($n_{gw}=\frac{n_g}{n_w}$).

Calculating the New Focal Length

Let's set up a ratio to compare the focal length in water ($f_{water}$) to the focal length in air ($f_{air}$). By dividing the Lens Maker's Formula for air by the formula for water, the geometric terms cancel out perfectly:

We know the refractive index of glass is $n_g=\frac{3}{2}$ and for water is $n_w=\frac{4}{3}$. Let's substitute these values carefully:

Simplifying the numerator and the denominator:

This is a profound result! The new focal length is four times the original focal length: $f_{water}=4f_{air}$.

Why the Image Disappears

Because the focal length has increased so drastically, the lens has lost a significant amount of its converging power. The light rays are not bent as sharply as they were in the air.

Consequently, the rays will no longer converge at the screen located at $2f_{air}$. In fact, because the object is now located at a distance ($u=2f_{air}$) which is less than the new focal length ($f_{water}=4f_{air}$), the rays will never converge on the right side of the lens at all! Therefore, the image on the screen simply disappears.

Where Did the Image Go?

If the image isn't on the screen, where is it? We can find out by applying the standard lens formula with our new parameters: $u=-2f_{air}$ and $f=+4f_{air}$.

Solving for $v$, we get $v=-4f_{air}$.

The negative sign indicates that a virtual, erect, and magnified image is formed on the same side as the object, at a distance of $4f_{air}$ from the lens. The screen remains blank, but a whole new virtual world has appeared behind the lens!