Rays from the sun converge at a point $25\text{ cm}$ behind a convex lens. The distance at which an object be placed in front of the lens to get a virtual image, is:

Rays coming from the sun are effectively parallel rays from an object at infinity. As per the chapter (Fig. 9.12a, Activity 9.11), 'several rays of light parallel to the principal axis... after refraction from the lens, are converging to a point on the principal axis. This point... is called the principal focus.' Since these rays converge $25\text{ cm}$ behind the convex lens, that point is the principal focus $F_2$, so the focal length is $f=+25\text{ cm}$ (positive by the New Cartesian Sign Convention, since it is a convex lens).

From Table 9.4 of the chapter, a convex lens forms a virtual, erect, magnified image only when the object lies between the optical centre $O$ and the focus $F_1$, i.e. when the object distance is numerically less than the focal length ($0<|u|<f$). For every other position ($u=f$, or $|u|>f$) the image is real.

Checking the options against $f = 25\text{ cm}$:

• (A) $20\text{ cm}$: since $20\text{ cm} < 25\text{ cm}$, the object lies between $O$ and $F_1$ $\rightarrow$ a virtual image is formed. This is the required condition, so (A) is correct.
• (B) $40\text{ cm}$: lies beyond $F_1$ ($40 > 25$), so a real, inverted image is formed — not virtual. Incorrect.
• (C) $50\text{ cm}$: this equals $2f$, i.e. the object is at $2F_1$; the image forms at $2F_2$, same size, real and inverted (Table 9.4). Incorrect.
• (D) More than $50\text{ cm}$ (beyond $2F_1$): image forms between $F_2$ and $2F_2$, diminished, real and inverted. Incorrect.

Verification using the lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ with $u=-20\text{ cm}$, $f=+25\text{ cm}$: $\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{25}-\dfrac{1}{20}=\dfrac{4-5}{100}=-\dfrac{1}{100}$, so $v=-100\text{ cm}$ — negative $v$ (same side as the object) confirms a virtual image, consistent with option (A).