Given data

• First convex lens: $f_1 = 30\,\text{cm} = 0.30\,\text{m}$
• Second convex lens: $f_2 = 10\,\text{cm} = 0.10\,\text{m}$
• Separation between lenses: $d = 10\,\text{cm} = 0.10\,\text{m}$

Power of each lens $P_1 = \frac{1}{f_1} = \frac{1}{0.30} = \frac{10}{3}\,\text{D}$ $P_2 = \frac{1}{f_2} = \frac{1}{0.10} = 10\,\text{D}$

Equivalent power of two thin lenses separated by distance $d$ For two thin lenses in air separated by distance $d$, $P_{\text{eq}} = P_1 + P_2 - d P_1 P_2$ where $d$ is in meters.

Substitute the values $P_{\text{eq}} = \frac{10}{3} + 10 - (0.10)\left(\frac{10}{3}\right)(10)$

Calculate the third term: $0.10 \cdot \frac{10}{3} \cdot 10 = \frac{10}{3}$

So, $P_{\text{eq}} = \frac{10}{3} + 10 - \frac{10}{3} = 10\,\text{D}$

Match with options

• A: $5\,\text{D}$
• B: $1\,\text{D}$
• C: $20\,\text{D}$
• D: $10\,\text{D}$

Hence, the correct option is D.

Comparison with stored correct answer Stored correct answer = D

This matches our derived answer.