(a) We know for a concave mirror f < 0 [negative] and u < 0 [negative] 2f < u < f ∴ $\frac{1}{2f}$ > $\frac{1}{u}$ > $\frac{1}{f}$ or $-\frac{1}{2f}$ < $\frac{-1}{u}$ < $\frac{-1}{f}$ or $\frac{1}{f} - \frac{1}{2f}$ < $\frac{1}{f} - \frac{1}{u}$ < $\frac{1}{f} - \frac{1}{f}$ or $\frac{1}{2f}$ < $\frac{1}{v}$ < 0 [Since $\frac{1}{f} - \frac{1}{u} = \frac{1}{v}$] Which implies that v < 0 to form image on the left. Also 2f > v {Since 2f and v are –ve} |2f| < |v| So, the real image is formed beyond 2f. (b) For a convex mirror, f > 0, always positive and object distance u < 0, always negative. Now mirror formula, $\frac{1}{f}$ = $\frac{1}{v} + \frac{1}{u}$ or $\frac{1}{v}$ = $\frac{1}{f} - \frac{1}{u}$ This implies that $\frac{1}{v}$ > 0 , or v > 0. So, whatever be the value of u, a convex mirror always forms a virtual image. (c) In convex mirror focal length is positive hence f > 0 and for an object distance from mirror with negative sign (u < 0) So, $\frac{1}{v} + \frac{1}{u}$ = $\frac{1}{f}$ or $\frac{1}{v}$ = $\frac{1}{f} - \frac{1}{u}$ The results $\frac{1}{v}$ > $\frac{1}{f}$ or v < f (both positive) hence the image is located between pole and focus of the mirror. Also magnification m = - $\frac{v}{u}$ = - $\frac{+v}{-u}$ m < [1] (positive) So, the image is virtual and diminished. (d) In concave mirror, f < 0 for object placed between focus and pole of concave mirror f < u < 0 (both negative) $\frac{1}{f}$ > $\frac{1}{u}$ Now mirror formula, $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ $\frac{1}{v}$ > 0 or v > 0 (positive) hence the image is virtual. Also magnification m = - $\frac{v}{u}$ , here $\frac{1}{v}$ < $\frac{1}{|u|}$ or v > |u| So, m > 1 hence the image is enlarged.