A beam of light passing through a diverging lens of focal length 0.3 m appears to focus 0.2 m behind the lens (so $v = -0.2$ m). Where is the object located?

1. For a diverging lens $f = -0.3$ m, and here $v = -0.2$ m. 2. Lens formula rearranged: $\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f}$. 3. Substitute: $\dfrac{1}{u} = \dfrac{1}{-0.2} - \dfrac{1}{-0.3} = -5 + 3.333 = -1.667$. 4. So $u = \dfrac{1}{-1.667} = -0.6$ m. 5. The negative sign places the object 0.6 m in front of the lens. 6. Choosing 0.2 m or 0.3 m simply repeats $v$ or $f$, which is the common sign-handling trap. _Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 2 Optics "Solved Problems", p.35_