1. Calculate Refractive Index ($\mu$):
$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$.

2. Determine Radius of Curvature ($R$):
For a lens with radius of aperture $r = 3 \text{ cm}$ and thickness $t = 0.3 \text{ cm}$, the radius of curvature is given by $R \approx \frac{r^2}{2t}$ (for small thickness).
$R = \frac{3^2}{2 \times 0.3} = \frac{9}{0.6} = 15 \text{ cm}$.

3. Lens Maker's Formula:
$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. For a plano-convex lens, $R_1 = R$ and $R_2 = \infty$.
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{15} \right) = 0.5 \times \frac{1}{15} = \frac{1}{30}$.
Thus, $f = 30 \text{ cm}$.

We need $0 < st < 1$, meaning $st$ is a small positive number. This requires $s$ and $t$ to have the same sign.

• Option A: $s < -1, t > 0$ → $st < 0$. Not possible.
• Option B: $s < -1, t < -1$ → both negative, so $st > 0$. For example $s = -2, t = -0.3$: $st = 0.6$, which satisfies $0 < st < 1$. ✓
• Option C: $s > -1, t < -1$ → could give $st$ negative or positive depending on sign of $s$. If $s = 0.5, t = -2$: $st = -1 < 0$. Not valid for all cases.
• Option D: $s > 1, t < -1$ → $st < -1 < 0$. Not possible.

Option B is the one that can satisfy the condition.

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