(c) $\frac{I_{1}}{I_{2}} = \frac{a^{2}}{b^{2}} = \frac{9}{1}$ $\therefore\ \ \ \frac{a}{b} = \frac{3}{1}$ $\frac{I_{\max}}{I_{\min}} = \frac{(a + b)^{2}}{(a - b)^{2}} = \left( \frac{3 + 1}{3 - 1} \right)^{2} = 4:1$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.