Question

Under what condition, a concave lens can make a real image?

• A. Real object placed between optical centre $O$ and and focus $F_2$.
• B. Virtual object placed between optical centre $O$ and focus $F_1$.
• C. Real object placed at focus $F_2$.
• D. Real object placed between $-\infty$ and $2F_2$.

Answer 1

The correct option is B Virtual object placed between optical centre $O$ and focus $F_1$.

Applying lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
We will be substituting only sign convention of $f$ and based on imposed condition $u$ will be obtained.
$f\to -f$
For a concave lens,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{-f}$
$\Rightarrow \frac{1}{v}=\frac{1}{u}-\frac{1}{f}$ ...$\left(i\right)$
For a real image, it must form along direction of incidence.
i.e. $v\to +ve$
Thus, for Eq. $\left(i\right)$ to be hold,
$\frac{1}{u}-\frac{1}{f}>0$
or, $u<f$ and $u$ must be positive.
$\therefore u$ should be between optical centre $O$ & $F_1$.

Why this question? It intends to challenge your fundamental understanding of sign convention of virtual object.