If the formation of a real image using a biconvex lens of material of refractive index 1.5 is shown in the figure. If this setup is immersed in water (refractive index = $\frac{4}{3}$), then
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To analyze changes in image formation when a lens is immersed in a medium, first calculate the new focal length using the lens maker's formula in the medium. Then, compare the object's position relative to this new focal length. If a real image previously formed, and the new focal length is significantly longer such that the object falls within it, the image will become virtual and disappear from any screen.
Step 1: Understand the Initial Setup and Given Information
The problem describes a biconvex lens with a refractive index $n_l = 1.5$.
Initially, this lens forms a real image. A real image can be projected onto a screen.
The diagram shows the formation of a real image when the lens is presumably in air.
Step 2: Identify the Change in Setup
The entire setup (lens and object) is immersed in water.
The refractive index of water is given as $n_w = \frac{4}{3}$.
Step 3: Determine the Effect of Immersion on Focal Length
When a lens is immersed in a medium, its focal length changes. The lens maker's formula is crucial here.
For a lens in air (refractive index $n_{air} \approx 1$):
\[
\frac{1}{f_{air}} = (n_l - n_{air}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]
For a lens immersed in a medium (water) with refractive index $n_m$:
\[
\frac{1}{f_{medium}} = \left( \frac{n_l}{n_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]
Let's find the ratio of focal lengths:
\[
\frac{f_{medium}}{f_{air}} = \frac{(n_l - 1)}{\left( \frac{n_l}{n_m} - 1 \right)}
\]
Step 4: Calculate the New Focal Length
Given values: $n_l = 1.5$, $n_m = n_w = \frac{4}{3}$.
Substitute these values into the ratio formula:
\[
\frac{f_{medium}}{f_{air}} = \frac{(1.5 - 1)}{\left( \frac{1.5 \times 3}{4} - 1 \right)}
\]
\[
\frac{f_{medium}}{f_{air}} = \frac{0.5}{\left( \frac{4.5}{4} - 1 \right)}
\]
\[
\frac{f_{medium}}{f_{air}} = \frac{0.5}{\left( 1.125 - 1 \right)}
\]
\[
\frac{f_{medium}}{f_{air}} = \frac{0.5}{0.125}
\]
\[
\frac{f_{medium}}{f_{air}} = 4
\]
So, the new focal length in water ($f_{medium}$) is 4 times the focal length in air ($f_{air}$).
\[
f_{medium} = 4 f_{air}
\]
Step 5: Analyze the Change in Image Formation
The original diagram shows the object placed at $2f_{air}$ (or very close to it, as a real image is formed). This means the object distance $u_{air}$ was such that $u_{air}>f_{air}$. In the specific diagram, the object is at $2f_{air}$, resulting in a real, inverted image of the same size at $2f_{air}$.
Now, the focal length has increased to $f_{medium} = 4f_{air}$.
The object distance ($u$) remains the same as it was initially with respect to the lens.
If the object was at $u = 2f_{air}$, then relative to the new focal length, the object distance is:
\[
u = 2f_{air} = \frac{1}{2} f_{medium}
\]
For a convex lens, if the object is placed between the optical center and the focal point ($u<f$), the image formed is virtual, erect, and magnified. In our case, $u = \frac{1}{2} f_{medium}$, which means the object is inside the new focal length.
Step 6: Conclude the Nature of the Image
Since the object is now placed within the focal length of the lens when immersed in water, the image formed will be virtual.
A virtual image cannot be projected onto a screen.
Step 7: Evaluate the Options
\begin{itemize}
\item Option (1): the image disappears from the screen. This is consistent with our conclusion that the image becomes virtual and cannot be projected.
\item Option (2): the image is magnified. While the virtual image would be magnified, the primary change is its nature (from real to virtual), which means it will no longer appear on the screen.
\item Option (3): the image will be real and erect. This is incorrect. A real image is always inverted.
\item Option (4): no change in the nature of the image. This is incorrect because the focal length changes, fundamentally altering the image formation.
\end{itemize}
Therefore, the most accurate description of what happens is that the image disappears from the screen.
