A thin convex lens has a power of $+8.0\text{D}$. An object is placed $30\text{cm}$ in front of the lens. Calculate the image distance $v$ from the lens. State whether the image is real or virtual and upright or inverted.

Step-by-step solution

1. First convert the power to focal length: $P=\frac{1}{f}$ (in metres), so $f=\frac{1}{P}=\frac{1}{8.0}=0.125\text{m}=12.5\text{cm}$.
2. Use the lens formula $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, where $u$ is the object distance (taken negative for a real object). Here $u=-30\text{cm}$.
3. Substitute the values: $\frac{1}{12.5}=\frac{1}{v}+\frac{1}{-30}$.
4. Rewrite: $\frac{1}{v}=\frac{1}{12.5}+\frac{1}{30}=0.08+0.0333=0.1133{\text{cm}}^{-1}$.
5. Hence $v=\frac{1}{0.1133}=8.83\text{cm}$. Wait, sign mistake: the object distance should be taken as $u=-30\text{cm}$, so $\frac{1}{v}=\frac{1}{12.5}+\frac{1}{30}$ is correct, giving $v=8.83\text{cm}$ on the opposite side of the lens, which is positive, indicating a real image.
6. But the calculation above missed a sign: the correct lens formula is $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ with $u=-30\text{cm}$, so $\frac{1}{v}=\frac{1}{12.5}-\frac{1}{30}=0.08-0.0333=0.0467{\text{cm}}^{-1}$.
7. Thus $v=\frac{1}{0.0467}=21.4\text{cm}$. Since $v$ is positive, the image forms on the opposite side of the lens, i.e., it is real and inverted.
8. Finally, round to appropriate significant figures: $v\approx 21.4\text{cm}$.
9. Therefore, the image distance is $21.4\text{cm}$, the image is real and inverted.

Concept tested: Lens formula and sign convention

Published 6 September 2026 · How this solution was prepared and checked