1. **State the problem:** The light intensity $I$ varies inversely with the square of the distance $d$ from the projector, so the relationship is given by: $$I = \frac{k}{d^2}$$ where $k$ is a constant. 2. **Given:** When $d = 3$ m, $I = 24$ units. 3. **Find the constant $k$:** Using the formula: $$24 = \frac{k}{3^2} = \frac{k}{9}$$ Multiply both sides by 9: $$k = 24 \times 9 = 216$$ 4. **a) Find the light intensity when $d = 6$ m:** Use the formula with $k=216$: $$I = \frac{216}{6^2} = \frac{216}{36} = 6$$ 5. **b) Effect when distance is halved:** If distance is halved, $d$ becomes $\frac{d}{2}$. Intensity becomes: $$I' = \frac{k}{(\frac{d}{2})^2} = \frac{k}{\frac{d^2}{4}} = 4 \times \frac{k}{d^2} = 4I$$ So, the intensity becomes 4 times greater. 6. **c) Effect when distance is increased by 25%:** New distance: $$d' = 1.25d$$ New intensity: $$I' = \frac{k}{(1.25d)^2} = \frac{k}{1.5625 d^2} = \frac{1}{1.5625} I = 0.64 I$$ Intensity decreases to 64% of original. 7. **Summary:** - At 6 m, intensity is 6 units. - Halving distance quadruples intensity. - Increasing distance by 25% reduces intensity to 64%.