Given data

• A convex lens forms a virtual image.
• Magnification is $3$.
• Distance between object and image is $20\,\text{cm}$.

Use magnification formula

For a lens, $m = \frac{v}{u}$

Since the image is virtual and magnified by $3$, $m = +3$ so, $\frac{v}{u} = 3 \implies v = 3u$

Here, for a convex lens producing a virtual image, both $u$ and $v$ are negative in sign convention. The relation $v=3u$ is still valid.

Use the given separation

Let the object be at distance $|u|$ from the lens and the virtual image be at distance $|v|$ from the lens on the same side.

Since both are on the same side of the lens, the distance between them is $|v| - |u| = 20$

But $|v| = 3|u|$, so $3|u| - |u| = 20$ $2|u| = 20$ $|u| = 10$

Hence, $u = -10\,\text{cm}, \qquad v = -30\,\text{cm}$

Apply lens formula

Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Substituting: $\frac{1}{f} = \frac{1}{-30} - \frac{1}{-10}$ $\frac{1}{f} = -\frac{1}{30} + \frac{1}{10}$ $\frac{1}{f} = \frac{-1+3}{30} = \frac{2}{30} = \frac{1}{15}$

Therefore, $f = 15\,\text{cm}$

Comparison with stored answer

Derived answer is $15$, which matches the stored correct answer.