1. **Problem statement:** We are asked to find an expression for the magnification $M$ of a lens and then determine if the image is upright or inverted when the object distance $x=3$ mm. 2. **Formula for magnification:** The magnification $M$ of a lens is given by the ratio of the image distance $v$ to the object distance $u$, with a negative sign: $$M = -\frac{v}{u}$$ This means: - If $M$ is positive, the image is upright. - If $M$ is negative, the image is inverted. 3. **Expression for $M$ using lens formula:** The lens formula relates object distance $u$, image distance $v$, and focal length $f$: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ Rearranging for $v$: $$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{u+f}{fu}$$ So, $$v = \frac{fu}{u+f}$$ Substitute $v$ into magnification formula: $$M = -\frac{v}{u} = -\frac{\frac{fu}{u+f}}{u} = -\frac{f}{u+f}$$ 4. **Final expression:** $$\boxed{M = -\frac{f}{u+f}}$$ 5. **Determine image orientation for $x=3$ mm:** Assuming $u = x = 3$ mm and a positive focal length $f$ (typical for converging lens), substitute into $M$: $$M = -\frac{f}{3 + f}$$ Since $f > 0$ and $3 + f > 0$, the fraction $\frac{f}{3+f}$ is positive, so $M$ is negative. 6. **Interpretation:** Because $M$ is negative, the image is inverted. **Answer:** - a) $M = -\frac{f}{u+f}$ - b) For $u=3$ mm, $M$ is negative, so the image is inverted. Image is inverted.