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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.

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(a) For concave mirror. f < 0, u < 0

∴ 2f < u < f ⇒ 1/2f > 1/u > 1/f

⇒ -1/2f < -1/u < -1/f

⇒ 1/f - 1/2f < 1/f - 1/u < 1/f - 1/f

∴ According to equation, 1/v = 1/f - 1/u

∴ v is negative. Therefore image formed is real and lies beyond 2f

(b) For convex mirror, f > 0, u < 0

As, 1/v = 1/f - 1/u

∴ v is positive. Therefore image formed is virtual and is at the back of the mirror.

(c) For convex mirror, f > 0, u < 0.

∴ 1/v = 1/f  - 1/u ∴ 1/v > 1/f or, v < f

therefore image is located between the pole and the focus. 

Also v < u, hence image is diminished.

(d) For concave mirror, f < 0, 

Since the object is placed between the pole and focus

∴ f < u < 0  ∴ (1/f - 1/u) > 0

Also, 1/f - 1/u = 1/v

∴ 1/v > 0  or v is positive, on the right side of mirror hence it must be virtual.

∴ v > u, ∴ image is enlarged.

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