An object of height $h$ is placed midway between $f$ and $2 f$ in front of a biconvex lens. A real inverted image is captured on a screen placed a little beyond $2 f$ on the other side of the lens. If the whole arrangement is immersed in water without disturbing the object, lens and the screen positions, then the new image formed will be

The focal length of a lens in a medium of refractive index $n_m$ is given by the lens maker formula: $\dfrac{1}{f_m} = (\dfrac{n_l}{n_m} - 1)(\dfrac{1}{R_1} - \dfrac{1}{R_2})$.

Initially, in air ($n_m = 1$), the focal length is $f$. When immersed in water ($n_m = 1.33$), the refractive index ratio $\dfrac{n_l}{n_m}$ decreases, which increases the focal length $f_m$ of the lens. Let the new focal length be $f' > f$.

The object is placed at a distance $u$ such that $f < u < 2f$. Since the lens is now in water, the new focal length $f'$ satisfies $f' > f$. The object position $u$ now satisfies $u < f'$.

When an object is placed within the focal length of a convex lens ($u < f'$), the lens acts as a simple magnifier. The image formed is virtual, erect, and magnified, and it is formed on the same side of the lens as the object.

Since the screen is fixed at the original position beyond $2f$, and the image is now virtual and formed on the same side as the object, no real image will be captured on the screen.

Answer: Magnified virtual erect image on the same side of the lens where the object is placed