(a) Object between f and 2f (Concave Mirror)

Given: f < u < 2f

For a concave mirror, focal length (f) and object distance (u) are negative. Let's work with magnitudes for simplicity:

f < u < 2f

Inverting the inequality (reverses signs):

1/f > 1/u > 1/2f

Multiplying by -1 (reverses signs again):

-1/f < -1/u < -1/2f

Using the Mirror Formula (1/v = 1/f - 1/u):

Substitute (-1/u) into the equation:

1/v < 1/f - 1/2f

1/v < 1/2f

Therefore, v > 2f

Since v is negative (real image) and its magnitude is greater than 2f, the image is formed beyond 2f.

(b) Convex Mirror (Always Virtual)

For a convex mirror:

• Focal length (f) is positive (+).
• Object distance (u) is negative (-).

Using the Mirror Formula:

1/v = 1/f - 1/u

Since u is negative, let u = -d (where d is distance):

1/v = 1/f + 1/d

Since both terms (1/f and 1/d) are positive, 1/v must be positive. This means v is always positive, which implies a Virtual Image formed behind the mirror.

(c) Convex Mirror (Diminished & Position)

Position:

From (b), we know 1/v = 1/f + 1/d.

This implies that 1/v > 1/f.

Therefore, v < f.

Since v is positive, the image is located between the Pole and Focus.

Size (Magnification):

m = -v / u

Since v is positive and u is negative, m is positive (erect image).

Since v < f and |u| is usually > f, the magnitude |m| < 1.

Therefore, the image is always diminished.

(d) Object between Pole and Focus (Concave Mirror)

Given: 0 < u < f (magnitudes)

In coordinates: -f < u < 0

Using the Mirror Formula:

1/v = 1/f - 1/u

Since 1/u is a larger negative number than 1/f (because |u| < |f|), the term -1/u becomes a large positive value that overpowers the negative 1/f.

Example: Let f = -10, u = -5.

1/v = -1/10 - (-1/5)

1/v = -1/10 + 2/10 = +1/10

Since 1/v is positive, v is positive (Virtual Image).

Also, magnification m = -v/u. Since v is positive and u is negative, m is positive (Erect). Since |v| > |u|, m > 1 (Enlarged).