9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.

2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

9.17 (a) sin i¢c = 1.44/1.68 which gives i¢c = 59°. Total internal reflection takes place when i > 59° or when r < rmax = 31°. Now, (sin i /sin r max max ) = 1.68 , which gives imax ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i¢c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i¢ = 53.5° which is greater than i¢c. Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections.