1. **State the problem:** We want to find the power $P$ of a magnifying glass lens that achieves an angular magnification $M=4$ when the object is placed 10 cm in front of the lens and the lens is held 8 cm from the eye. 2. **Given formula:** The problem provides the formula $$M = \frac{qL}{1 - dL'}$$ where: - $M$ is the angular magnification, - $q$ is the near point distance of the eye (usually 25 cm), - $L$ is the power of the lens, - $d$ is the distance from the lens to the eye, - $L'$ is the focal length reciprocal of the lens (related to $L$). 3. **Important notes:** Power $L$ of a lens is related to focal length $f$ by $$L = \frac{100}{f}$$ where $f$ is in cm and $L$ in diopters. 4. **Identify variables:** - $M = 4$ - $q = 25$ cm (standard near point) - $d = 8$ cm 5. **Rewrite the formula:** We want to find $L$ (power), and $L' = \frac{1}{f}$ (in cm$^{-1}$). Since $L = \frac{100}{f}$, then $$L' = \frac{1}{f} = \frac{L}{100}$$ 6. **Substitute $L'$ into the formula:** $$M = \frac{qL}{1 - d \frac{L}{100}} = \frac{qL}{1 - \frac{dL}{100}}$$ 7. **Plug in known values:** $$4 = \frac{25L}{1 - \frac{8L}{100}} = \frac{25L}{1 - 0.08L}$$ 8. **Solve for $L$:** Multiply both sides by denominator: $$4(1 - 0.08L) = 25L$$ $$4 - 0.32L = 25L$$ $$4 = 25L + 0.32L = 25.32L$$ $$L = \frac{4}{25.32} \approx 0.158$$ diopters 9. **Interpretation:** The power of the lens needed is approximately $0.158$ diopters. **Final answer:** $$\boxed{L \approx 0.16 \text{ diopters}}$$