Use the mirror equation to deduce that: (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
For a concave mirror, f is negative. When object is placed between f and 2f, we have 2f < u < f (both negative). Using mirror equation: 1/v = 1/f - 1/u. Since |u| lies between |f| and 2|f|, we get 1/v = 1/f - 1/u. Substituting f = -f₀ and u = -u₀ (where f₀ > 0, u₀ > 0): 1/v = -1/f₀ + 1/u₀ = (u₀ - f₀)/(f₀u₀). Since u₀ > f₀ (object beyond focus), we get 1/v < 0, so v < 0 (real image). Also, |v| = f₀u₀/(u₀ - f₀). Since u₀ < 2f₀, we can show |v| > 2f₀. Thus, image is real and beyond 2f.
For a convex mirror, f is positive. Object distance u is always negative. From mirror equation: 1/v = 1/f - 1/u. Since f > 0 and u < 0, we get 1/v = 1/f + 1/|u|, which is always positive. Hence v > 0 always, meaning the image is always virtual regardless of object position.
For convex mirror with f > 0 and u < 0: From 1/v = 1/f + 1/|u|, we get v = f|u|/(|u| + f). Since |u| > 0, we have v < f, meaning image lies between pole and focus. Magnification m = -v/u = v/|u| = f/(|u| + f). Since |u| > 0, we get m < 1, meaning image is always diminished. Also m > 0 indicates erect image.
For concave mirror with f < 0, when object is between pole and focus: 0 < |u| < |f|. Let f = -f₀ and u = -u₀ where u₀ < f₀. From mirror equation: 1/v = (f₀ - u₀)/(f₀u₀). Since u₀ < f₀, we get 1/v > 0, so v > 0 (virtual image). Magnification m = -v/u = f₀/(f₀ - u₀). Since u₀ < f₀, denominator is less than f₀, so m > 1 (enlarged). Also m > 0 (erect image).
Explanation
This question requires algebraic deduction using the mirror equation (1/v + 1/u = 1/f) and magnification formula (m = -v/u) with proper sign convention. For concave mirrors, f is negative; for convex mirrors, f is positive. Object distance u is always negative. Real images have v < 0, virtual images have v > 0. Each sub-part can be proven by substituting the given conditions into the mirror equation and analyzing the sign and magnitude of v and m.